Math 3 Quiz: Verifying Solutions With Technology

20 questions · exam conditions

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Verifying Solutions With TechnologyQuestion 1 of 20

A student uses technology to find that $\log_2(x) + \log_2(x-3) = 2$ has solution $x = 4$ . When verifying: $\log_2(4) + \log_2(1) = 2 + 0 = 2$ ✓. However, the technology also suggested $x ≈ 3.99$ as an approximate solution. When the student checks this: $\log_2(3.99) + \log_2(0.99) ≈ 1.996 + (-0.014) ≈ 1.982$ . Why does this approximation perform worse than the exact value?

Logarithmic functions are inherently unstable near their exact solutions

The equation actually has two solutions, and 3.99 corresponds to the second solution

Small errors in x create larger relative errors when x-3 is close to zero

The calculator's logarithm function loses accuracy for values very close to integers

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Math 3 Quiz

Math 3 Quiz: Verifying Solutions With Technology

Practice Verifying Solutions With Technology in Math 3 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Verifying Solutions With Technology, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Logarithmic functions are inherently unstable near their exact solutions
The equation actually has two solutions, and 3.99 corresponds to the second solution
Small errors in x create larger relative errors when x-3 is close to zero (correct answer)
The calculator's logarithm function loses accuracy for values very close to integers

Explanation: The correct answer is C. When x = 4, we have x - 3 = 1, but when x = 3.99, we have x - 3 = 0.99. The small change in x creates a proportionally larger change in (x-3) when x-3 is small, and logarithms amplify these differences. This demonstrates how verification can reveal that some approximate solutions are less reliable than exact ones.

Question 2

A student uses a graphing calculator to solve the equation $2^{x-1} = 3x - 1$ and finds the solution $x ≈ 1.17$ . When they substitute this value back into the original equation, they get $2^{1.17-1} ≈ 2.25$ and $3(1.17) - 1 = 2.51$ . What is the most likely explanation for this discrepancy?

The equation has no real solutions, so the calculator result is invalid
The calculator rounded the solution, causing accumulated error in verification (correct answer)
The student made an algebraic error when setting up the original equation
The exponential function was entered incorrectly into the calculator initially

Explanation: The discrepancy between 2.25 and 2.51 is due to rounding error. The calculator likely found a more precise solution (perhaps x ≈ 1.1739...) but displayed only two decimal places. When the rounded value is substituted back, it doesn't satisfy the equation exactly. This is a common issue when verifying solutions with technology - the displayed rounded values may not appear to satisfy the original equation perfectly.

Question 3

A student uses technology to solve $2^x + 2^{-x} = 3$ and finds $x ≈ 1.585$ and $x ≈ -1.585$ . When verifying $x = 1.585$ , they calculate $2^{1.585} ≈ 3.00$ and $2^{-1.585} ≈ 0.333$ , giving a sum of approximately $3.333$ . What is the most likely cause of this discrepancy?

The equation has no real solutions, so the technology results are computational artifacts
The verification process should use logarithmic properties rather than direct substitution
Exponential equations with negative exponents require specialized numerical methods for accuracy
The student used insufficient precision when reading the solution from the calculator display (correct answer)

Explanation: When solving exponential equations with technology, precision in reading and using solutions is crucial. Let's examine what happened here. The correct answer is D because the student clearly rounded the solution when copying it from their calculator. When you substitute

x = 1.585

back into the original equation, you get approximately 3.333 instead of 3, indicating the value isn't precise enough. If the student had used more decimal places (like

x ≈ 1.58496...

), the verification would have been much closer to 3. Let's see why the other options miss the mark: A is incorrect because this equation definitely has real solutions. You can verify this algebraically by substituting

y = 2^x

to get

y + \frac{1}{y} = 3

, which leads to

y^2 - 3y + 1 = 0

. This quadratic has real solutions. B is wrong because direct substitution is the standard and correct way to verify solutions. Logarithmic properties aren't needed for verification—they might be used in solving, but substitution works perfectly fine here. C is incorrect because negative exponents don't require specialized methods. Standard calculators handle

2^{-x}

just fine using the relationship

2^{-x} = \frac{1}{2^x}

. Study tip: When using technology to solve equations, always copy more decimal places than you think you need for verification. Rounding too early is one of the most common sources of error in numerical problem-solving. Your calculator likely displays more precision than what initially appears on screen.

Question 4

A student graphs the function $f(x) = \frac{x^2 - 4}{x - 2}$ on a graphing calculator to find where $f(x) = 5$ . The graph appears to show a solution at $x = 2$ , but when they substitute $x = 2$ into the original equation $\frac{x^2 - 4}{x - 2} = 5$ , they get $\frac{0}{0}$ . How should they interpret this result?

The calculator correctly identified x = 2 as a solution despite the undefined expression
The function has a removable discontinuity at x = 2, so this cannot be a solution (correct answer)
They should zoom in on the graph to get a more accurate reading of the intersection
The equation $\frac{x^2 - 4}{x - 2} = 5$ should be solved algebraically instead of graphically

Explanation: The function f(x) = (x²-4)/(x-2) = (x+2)(x-2)/(x-2) simplifies to f(x) = x+2 for x ≠ 2, with a removable discontinuity (hole) at x = 2. The graph shows the line y = x+2 with a hole at (2,4). Since f(2) is undefined, x = 2 cannot be a solution to f(x) = 5. The student likely misread the graph or the calculator's resolution made the hole appear connected.

Question 5

A student uses a CAS (Computer Algebra System) to solve $\log_2(x) + \log_2(x-3) = 2$ and gets $x = 4$ . They verify by calculating $\log_2(4) + \log_2(4-3) = 2 + 0 = 2$ . However, when they graph $y = \log_2(x) + \log_2(x-3)$ and $y = 2$ , the intersection appears to be at $x ≈ 4.24$ . What explains this discrepancy?

The CAS solution is exact while the graphical intersection shows rounding error (correct answer)
The student made an error in the verification calculation for x = 4
The logarithm properties were applied incorrectly in the original equation
The graphing utility has difficulty with logarithmic functions near asymptotes

Explanation: The algebraic solution x = 4 is correct: log₂(4) + log₂(1) = 2 + 0 = 2. The graphical intersection appearing at x ≈ 4.24 is due to the limited resolution of the graphing utility and possibly the viewing window. The CAS provides the exact analytical solution, while the graph shows an approximation. This demonstrates why verification through substitution is often more reliable than reading coordinates from a graph.

Question 6

Using technology to solve $\sqrt{x + 3} = x - 1$ , a student finds $x = 6$ and $x = 1$ . Verification shows: For $x = 6$ : $\sqrt{6 + 3} = 3$ and $6 - 1 = 5$ . For $x = 1$ : $\sqrt{1 + 3} = 2$ and $1 - 1 = 0$ . What conclusion should the student draw about the technology's solutions?

Both solutions are invalid due to calculation errors in the verification process
The technology solved a different equation than the one originally entered
Neither solution satisfies the original equation, indicating the technology found extraneous solutions (correct answer)
Only x = 6 is valid since it produces values closer in magnitude than x = 1

Explanation: Both x = 6 and x = 1 fail verification in the original equation √(x+3) = x-1. The verification shows that neither value makes both sides equal. This commonly occurs when technology solves by squaring both sides, which can introduce extraneous solutions. The algebraic approach: squaring gives x+3 = (x-1)², leading to x²-3x-2 = 0. While this quadratic has solutions, they must be checked in the original equation where the square root requires the right side to be non-negative.

Question 7

Using a graphing calculator to find the maximum of $f(x) = -2x^2 + 8x - 5$ on the interval $[0, 5]$ , a student finds the maximum occurs at $x = 2.01$ with $f(2.01) = 2.96$ . Algebraically, they know the vertex is at $x = 2$ with $f(2) = 3$ . Which statement best explains this discrepancy?

The calculator's numerical methods have inherent precision limitations for optimization problems (correct answer)
The student incorrectly calculated the vertex formula, leading to the wrong analytical answer
The interval restriction [0, 5] affects the calculator's ability to find the true maximum
Quadratic functions require symbolic computation rather than numerical approximation methods

Explanation: The analytical solution is exact: the vertex occurs at x = -b/(2a) = -8/(2(-2)) = 2, giving f(2) = 3. The calculator's result of x = 2.01 with f(2.01) = 2.96 demonstrates the limitations of numerical optimization algorithms, which use iterative approximation methods. These methods converge to the solution but may not achieve perfect precision due to rounding errors in the computational process.

Question 8

A physics student models the height of a projectile with the equation $h(t) = -16t^2 + 64t + 80$ , where $h$ is height in feet and $t$ is time in seconds.

Using a graphing calculator to find when the projectile hits the ground ( $h = 0$ ), the student finds $t ≈ 5.099$ . When they verify by substitution, they calculate $h(5.099) = -16(5.099)^2 + 64(5.099) + 80 ≈ -16(25.999) + 326.336 + 80 ≈ 0.738$ . How should they interpret this non-zero result?

The projectile never actually hits the ground according to this mathematical model
The calculator's solution has adequate precision for the physical context of the problem
They should use the quadratic formula to find the exact algebraic solution instead
The rounding in intermediate calculations created cumulative error requiring more precision (correct answer)

Explanation: The result h(5.099) ≈ 0.738 feet indicates that the rounded value t ≈ 5.099 is not precise enough. The cumulative rounding error from squaring 5.099 and subsequent calculations amplifies the imprecision. For projectile problems, being off by nearly a foot suggests the need for more decimal places from the calculator's stored solution. The exact solution using the quadratic formula would be t = (4 + √21)/2 ≈ 5.09902...

Question 9

A student uses technology to find where $\frac{1}{x-1} = \frac{2x}{x+1}$ . The calculator suggests $x ≈ 2.414$ as a solution. Verification gives $\frac{1}{2.414-1} = \frac{1}{1.414} ≈ 0.707$ and $\frac{2(2.414)}{2.414+1} = \frac{4.828}{3.414} ≈ 1.414$ . What should the student conclude?

The calculator solution is incorrect and should be rejected completely
More decimal places are needed to verify this solution accurately (correct answer)
The equation has no real solutions, explaining the verification failure
Cross-multiplication should be used to avoid fractional calculation errors

Explanation: The significant discrepancy between 0.707 and 1.414 (factor of 2 difference) strongly suggests that x ≈ 2.414 is not precise enough. Given that 1.414 ≈ √2 and 0.707 ≈ 1/√2, the actual solution might be exactly √2 + 1 ≈ 2.41421356... The student needs more decimal places from the calculator's stored value to verify properly. The large discrepancy indicates insufficient precision rather than an incorrect solution method.

Question 10

Using technology to solve the system $\begin{cases} x^2 + y^2 = 25 \\ xy = 12 \end{cases}$ , a student finds four solutions including $(3, 4)$ . When they verify this solution, they calculate $3^2 + 4^2 = 9 + 16 = 25$ ✓ and $3 \cdot 4 = 12$ ✓. However, the graphing utility shows the intersection points slightly off the grid. What does this indicate?

The algebraic verification confirms the solution is exact despite graphical imprecision (correct answer)
The student should re-examine their manual calculations for potential arithmetic errors
The system has approximate solutions only, making exact verification impossible
The graphing utility's resolution is insufficient for accurately displaying integer coordinates

Explanation: The solution (3, 4) is exact, as confirmed by perfect algebraic verification. The fact that the graphical intersections appear slightly off-grid is due to the limited pixel resolution of the graphing display and potentially the scaling of the viewing window. This demonstrates that algebraic verification is more reliable than visual inspection of graphs for confirming exact solutions.

Question 11

A student solves $3\sin(x) = 2\cos(x)$ using technology and finds solutions $x ≈ 0.588$ and $x ≈ 3.730$ in the interval $[0, 2\pi]$ . When verifying $x = 0.588$ , they calculate $3\sin(0.588) ≈ 1.66$ and $2\cos(0.588) ≈ 1.67$ . The small difference suggests what about the solution?

The solution is sufficiently accurate given the inherent limitations of decimal approximations (correct answer)
The trigonometric functions require degree mode instead of radian mode for proper verification
A more precise value from the calculator's memory should be used for verification
The equation should be converted to $\tan(x) = \frac{2}{3}$ before solving numerically

Explanation: The close agreement between 1.66 and 1.67 (difference of 0.01) indicates that x ≈ 0.588 is a reasonably accurate solution. For trigonometric equations solved numerically, perfect agreement in verification is rarely achieved due to rounding in both the solution-finding process and the verification calculations. This level of accuracy is typically acceptable for practical purposes.

Question 12

A student graphs $y = \sqrt{x}$ and $y = 4 - x$ to find their intersection points. The technology shows intersection at $(4, 0)$ , but when they verify, they get $\sqrt{4} = 2$ and $4 - 4 = 0$ . Since $2 \neq 0$ , they conclude there's an error. Examining the graph more closely shows the curves don't actually intersect at $(4, 0)$ . What likely happened?

The graphing utility incorrectly calculated the intersection due to software limitations
The resolution settings need adjustment to show the true intersection point accurately
The square root function domain restriction caused an invalid intersection calculation
The student misread the coordinates of the intersection point from the graph display (correct answer)

Explanation: When working with graphing technology to find intersection points, always verify your results algebraically. The key insight here is recognizing when visual interpretation can lead to errors. To find where

y = \sqrt{x}

and

y = 4 - x

actually intersect, you set them equal:

\sqrt{x} = 4 - x

. Squaring both sides gives

x = (4-x)^2 = 16 - 8x + x^2

, which rearranges to

x^2 - 9x + 16 = 0

. Using the quadratic formula yields

x = \frac{9 \pm \sqrt{17}}{2}

, giving approximately

x = 2.44

and

x = 6.56

. However, when

x = 6.56

, we get

\sqrt{6.56} \approx 2.56

but

4 - 6.56 = -2.56

. Since we need

\sqrt{x} = 4 - x

and square roots are always non-negative, only

x \approx 2.44

gives a valid intersection at approximately

(2.44, 1.56)

. The correct answer is D because the student likely misread the graph coordinates, seeing

(4,0)

when the actual intersection is nowhere near that point. Choice A incorrectly blames software error when the technology likely displayed the correct intersection. Choice B suggests a resolution problem, but this is a misreading issue, not a display clarity problem. Choice C wrongly implies domain restrictions caused a calculation error, when the real issue is coordinate misinterpretation. Always double-check graphing technology results with algebraic verification, especially when the visual intersection seems unclear or when your verification doesn't match what you think you see.

Question 13

Using a graphing calculator to solve $x^3 - 6x^2 + 9x = 0$ , a student identifies three solutions: $x = 0$ , $x = 3$ , and $x = 3$ . When they substitute to verify, all values work perfectly: $0^3 - 6(0)^2 + 9(0) = 0$ , and $3^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0$ . However, the graph only shows two x-intercepts. What explains this apparent contradiction?

The calculator incorrectly identified a repeated root that doesn't actually exist
One of the x-intercepts represents a double root, so x = 3 has multiplicity 2 (correct answer)
The graphing window needs to be adjusted to show the third intersection point
Cubic equations can have complex roots that don't appear on real coordinate graphs

Explanation: Factoring the equation: x³ - 6x² + 9x = x(x² - 6x + 9) = x(x - 3)² = 0. This gives x = 0 (multiplicity 1) and x = 3 (multiplicity 2). The graph shows only two x-intercepts because the repeated root at x = 3 appears as a single point where the curve touches the x-axis tangentially rather than crossing it. Both solutions verify correctly, but the double root at x = 3 explains why there are three algebraic solutions but only two visible intercepts.

Question 14

Using a graphing utility, Maria finds that the system $\begin{cases} y = \ln(x + 2) \\ y = -0.5x + 1.8 \end{cases}$ has solutions at approximately $x = 0.89$ and $x = 2.31$ . When she checks $x = 2.31$ by substitution, she calculates $\ln(2.31 + 2) = \ln(4.31) ≈ 1.46$ and $-0.5(2.31) + 1.8 = 0.645$ . What should Maria conclude?

The graphing utility malfunctioned and provided an incorrect intersection point
She should use more decimal places in her verification calculations (correct answer)
The system actually has only one solution, not two as initially found
The logarithmic function domain restriction makes this solution invalid

Explanation: The large discrepancy between 1.46 and 0.645 suggests that x = 2.31 is not accurate enough. The graphing utility likely found a more precise solution, but Maria is using a rounded approximation. She should either use more decimal places from the calculator's stored value or re-examine the graph for a more precise intersection point. The solution exists since both functions are defined at x = 2.31, but more precision is needed for verification.

Question 15

A graphing calculator shows that the equation $e^x = 4 - x^2$ has two intersection points at approximately $x = -1.86$ and $x = 1.18$ . When a student verifies $x = 1.18$ , they calculate $e^{1.18} ≈ 3.25$ and $4 - (1.18)^2 ≈ 2.61$ . The difference is $0.64$ . What is the best next step?

Accept the solution since exponential equations typically have approximate solutions only
Conclude that the graphing window was set incorrectly, causing false intersections
Use the calculator's stored value with more precision for a better verification (correct answer)
Check whether the equation was entered correctly by re-graphing both functions

Explanation: A difference of 0.64 between the two sides suggests significant rounding error. The calculator likely found a more precise solution (perhaps x ≈ 1.1785...) but displayed only 2-3 decimal places. Using the stored value with full precision should yield much closer agreement. The intersection clearly exists graphically, so the issue is precision in verification, not an incorrect equation or false intersection.

Question 16

A student uses a graphing calculator to solve the equation $2^x = x + 3$ and finds the solution $x ≈ 2.54$ . When checking this value by substituting back into the original equation, the student calculates $2^{2.54} ≈ 5.82$ and $2.54 + 3 = 5.54$ . What best explains this discrepancy?

The calculator made an error in finding the intersection point of the two functions
The discrepancy is due to rounding the solution to two decimal places before verification (correct answer)
The student should have used logarithms instead of graphing to solve the equation
The equation has no real solution, so any approximation will show discrepancies

Explanation: The correct answer is B. When the calculator finds the intersection, it uses many more decimal places internally than what is displayed. By rounding to 2.54 and then substituting back, precision is lost. The actual solution is approximately 2.5440307..., and using this more precise value would show much closer agreement between the two sides of the equation.

Question 17

A student graphs $f(x) = \sqrt{x-1} + 2$ and $g(x) = 0.5x + 1.5$ on a calculator and identifies their intersection at approximately $(5, 4)$ . However, when verifying by substitution: $f(5) = \sqrt{5-1} + 2 = 4$ exactly, but $g(5) = 0.5(5) + 1.5 = 4$ exactly. The student is confused why the verification shows exact agreement. What explains this situation?

The calculator's trace function automatically rounds to exact integer coordinates when possible
This particular intersection point happens to have exact rational coordinates by coincidence (correct answer)
The student made calculation errors and the functions don't actually intersect at (5,4)
Graphing calculators are more accurate for radical functions than polynomial functions

Explanation: The correct answer is B. Sometimes intersection points do have exact rational coordinates. Setting

\sqrt{x-1} + 2 = 0.5x + 1.5

and solving algebraically:

\sqrt{x-1} = 0.5x - 0.5

, squaring both sides gives

x-1 = 0.25x^2 - 0.5x + 0.25

, which simplifies to

0.25x^2 - 1.5x + 1.25 = 0

. This yields

x = 1

x = 5

. The point

(5,4)

is indeed an exact solution, requiring no rounding.

Question 18

A calculator shows that the equation $\cos(x) = 0.6x + 0.1$ has a solution at $x ≈ 0.927$ . When a student verifies this by calculating $\cos(0.927) ≈ 0.596$ and $0.6(0.927) + 0.1 = 0.656$ , there's a noticeable difference. The student decides to use more precision and finds $x ≈ 0.9273$ . What verification result should they expect with this more precise value?

The discrepancy will completely disappear and both sides will be exactly equal
The trigonometric side will be more accurate but the linear side will be less accurate
The discrepancy will remain the same because the calculator has systematic error
The values will be much closer but still show a small discrepancy due to finite precision (correct answer)

Explanation: When you encounter equations that mix trigonometric and algebraic functions, solutions typically require numerical methods since they can't be solved algebraically. Understanding how precision affects verification is crucial for interpreting calculator results. The more precise value

x ≈ 0.9273

should bring the two sides much closer together, but won't eliminate the discrepancy entirely. This happens because calculators use finite decimal representations for numbers that may be irrational or have many decimal places. Even our "more precise" value of 0.9273 is still rounded. When you substitute this into both sides of

\cos(x) = 0.6x + 0.1

, you'll get values that are much closer than 0.596 and 0.656, but still won't match perfectly due to this inherent limitation of decimal approximation. Choice A is wrong because achieving exact equality would require infinite precision, which calculators cannot provide. Choice B incorrectly suggests that one side becomes less accurate—both sides benefit from the more precise x-value, though the trigonometric function

\cos(x)

tends to be more sensitive to input changes. Choice C misunderstands the source of error; this isn't systematic calculator error but rather the fundamental limitation of representing infinite-precision numbers with finite decimals. The linear expression

0.6x + 0.1

will be computed with high accuracy since it involves simple arithmetic, while

\cos(x)

involves more complex internal calculations but should still improve significantly. Remember: when verifying numerical solutions to transcendental equations, expect improvement with greater precision, but perfect matches only occur with exact solutions—which rarely exist for mixed function types.

Question 19

Using a graphing calculator's solver function, Alex finds that $3\sin(x) - x = 1$ has a solution at $x ≈ 2.029$ . When checking by substitution, $3\sin(2.029) ≈ 3(0.893) ≈ 2.679$ and $2.679 - 2.029 = 0.650$ , which is not close to 1. Alex increases the calculator's precision setting and gets $x ≈ 2.02876$ . What should Alex expect from this more precise value?

The verification will now show exact agreement because transcendental equations have exact solutions
Trigonometric equations cannot be verified accurately using decimal approximations
The discrepancy will remain large because the original equation was entered incorrectly
The result will be closer to 1, demonstrating the importance of sufficient precision in verification (correct answer)

Explanation: When you encounter transcendental equations like

3\sin(x) - x = 1

, you're dealing with equations that mix algebraic and trigonometric functions and typically can't be solved exactly by hand. These require numerical methods, and understanding precision is crucial for verification. Alex's situation perfectly demonstrates why precision matters in numerical solutions. With

x ≈ 2.029

, the verification gives

3\sin(2.029) - 2.029 ≈ 2.679 - 2.029 = 0.650

, which is far from the target value of 1. However, when using the more precise value

x ≈ 2.02876

, the result should be much closer to 1, confirming that the solver is working correctly but needs sufficient precision for accurate verification. Choice A is incorrect because transcendental equations rarely have exact rational solutions—they're typically irrational numbers that require approximation. Choice B is wrong because trigonometric equations can absolutely be verified with decimal approximations, provided you use enough precision. Choice C misses the point entirely—the equation was entered correctly, as evidenced by the solver finding a solution; the issue is simply precision in the approximation. The correct answer is D because increasing precision in numerical solutions directly improves verification accuracy. The more precise value will yield a result much closer to 1 when substituted back into the original equation. Study tip: When verifying solutions to transcendental equations, always use the full precision your calculator provides. Round only your final answer, never the intermediate values used for verification.

Question 20

A student solves $e^x - 3x = 2$ using a graphing calculator and obtains $x ≈ 1.857$ . During verification, they calculate $e^{1.857} ≈ 6.404$ and $3(1.857) = 5.571$ , giving $6.404 - 5.571 = 0.833$ , which is significantly different from 2. Before concluding the calculator is wrong, what should the student check first?

Whether 1.857 was copied correctly from the calculator display (correct answer)
Whether the calculator is in the correct angle mode for exponential functions
Whether the exponential function key is functioning properly on the calculator
Whether the original equation was graphed correctly as two separate functions

Explanation: When solving equations with technology, verification discrepancies often point to simple human errors rather than calculator malfunctions. Let's examine what's happening here. The student's verification reveals a significant error: they got 0.833 instead of 2. However, before questioning the calculator's accuracy, you should systematically check the most likely sources of error, starting with the simplest. Choice A is correct because transcription errors are extremely common when copying decimal values from calculator displays. Even a small error in copying—like writing 1.857 instead of the actual solution—would cause exactly this type of verification failure. This is the most probable explanation and easiest to check. Choice B is incorrect because angle modes (radians vs. degrees) only affect trigonometric functions, not exponential functions like

e^x

. The exponential function works the same regardless of angle mode setting. Choice C is incorrect because if the exponential function key were malfunctioning, the student wouldn't have been able to calculate

e^{1.857} ≈ 6.404

during verification. The fact that this calculation worked properly indicates the exponential function is operating correctly. Choice D is incorrect because the verification process itself—substituting the alleged solution back into the original equation—doesn't depend on how the equation was originally graphed. The graphing method is irrelevant to checking whether

e^x - 3x = 2

when

x = 1.857

. Study tip: When verification fails dramatically, always check transcription errors first—they're far more common than calculator malfunctions and much easier to fix.