AP Calculus AB

AP Calculus AB Practice Test: Practice Test 7

Practice Test 7 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A biologist observes that a population of yeast cells is growing in a nutrient broth.

To determine the population's growth rate at the exact moment it reaches 50,000 cells, what is the most appropriate calculus-based approach?

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Question 1

A biologist observes that a population of yeast cells is growing in a nutrient broth.

To determine the population's growth rate at the exact moment it reaches 50,000 cells, what is the most appropriate calculus-based approach?

Measure the population one hour before and one hour after it reaches 50,000 and find the average rate.
Divide 50,000 cells by the total time it took for the population to reach that size from the start.
Assume the growth rate is constant and equal to the average rate observed during the first hour of growth.
Calculate average growth rates over increasingly small time intervals around that moment and find their limiting value. (correct answer)

Explanation: The instantaneous growth rate cannot be measured directly. It must be determined by the limiting process of calculus. This involves finding the trend of average growth rates as the time interval for the measurement is made infinitesimally small around the moment of interest.

Question 2

Let $A(x)=\ln\!\big((x-2)^2+5\big)$ . What is $A'(x)$ ?

$\dfrac{1}{(x-2)^2+5}$
$\dfrac{2(x-2)}{(x-2)^2+5}$ (correct answer)
$\dfrac{2x}{(x-2)^2+5}$
$\dfrac{(x-2)^2+5}{2(x-2)}$
$2(x-2)\ln((x-2)^2+5)$

Explanation: A(x) = $\ln((x - 2)^2 + 5)$ , outer $\ln(u)$ , $u = (x - 2)^2 + 5$ . Derivative: $\frac{1}{u} \cdot 2(x - 2)$ . Inner quadratic. Omission: Missing $2(x - 2)$ . Some treat as $\ln(x)$ . Pattern: Logs of quadratics chain the quadratic derivative.

Question 3

A function $u$ satisfies $\lim_{x\to5}u(x)=7$ , but $u(5)$ is not defined. Is $u$ continuous at $5$ , and why?

Yes; the limit exists, so $u$ is continuous at $5$ .
No; $u(5)$ does not exist, so continuity fails even though $\lim_{x\to5}u(x)$ exists. (correct answer)
Yes; define $u(5)=0$ and it becomes continuous.
No; the limit does not exist because $u(5)$ is undefined.
Yes; $u(5)$ must equal $7$ automatically since the limit is $7$ .

Explanation: For continuity at x = 5, we need three conditions: (1) u(5) exists, (2) lim[x→5] u(x) exists, and (3) lim[x→5] u(x) = u(5). We're told that lim[x→5] u(x) = 7, so condition (2) is satisfied. However, we're also told that u(5) is not defined, which means condition (1) fails immediately. Without u(5) existing, there's no value to compare with the limit, so condition (3) cannot be satisfied either. Therefore, u is not continuous at x = 5. A common mistake is thinking that if the limit exists, we can just define the function value to make it continuous—but the question asks about the given function, not a modified version. Continuity checklist: ✗ Function defined at point, ✓ Limit exists, ✗ Limit equals function value.

Question 4

A bacteria culture grows at rate $B'(t)=6$ mg/hr for $0\le t\le1$ and $B'(t)=0$ for $1\le t\le4$ . Total change?

$24$ mg
$6$ mg (correct answer)
$0$ mg
$18$ mg
$4$ mg

Explanation: This problem uses accumulation reasoning to find the total change in bacteria mass from a piecewise constant rate function. The total change equals the sum of integrals over each interval with different growth rates. For 0 ≤ t ≤ 1: ∫6 dt = 6t evaluated from 0 to 1 gives 6 mg of growth. For 1 ≤ t ≤ 4: ∫0 dt = 0, meaning no additional growth occurs. The total change is 6 + 0 = 6 mg. Choice A (24 mg) might tempt students who multiply 6 mg/hr by the entire 4-hour period, not recognizing that growth stops after t = 1. For piecewise rate functions, integrate each piece only over its specified interval, as the accumulation depends on both the rate value and the duration.

Question 5

Given $\int_{5}^{8} s(x)\,dx=-3$ , what is $\int_{8}^{5} \left(\tfrac12 s(x)\right)\,dx$ ?

$-\tfrac{3}{2}$
$\tfrac{3}{2}$ (correct answer)
$-6$
$6$
$\tfrac{1}{6}$

Explanation: This problem combines the constant multiple rule and reversal property of definite integrals. Given $\int_{5}^{8} s(x)\,dx = -3$ , we need $\int_{8}^{5} \frac{1}{2}s(x)\,dx$ . By the reversal property, $\int_{8}^{5} \frac{1}{2}s(x)\,dx = -\int_{5}^{8} \frac{1}{2}s(x)\,dx$ . Using the constant multiple rule, $\int_{5}^{8} \frac{1}{2}s(x)\,dx = \frac{1}{2}\int_{5}^{8} s(x)\,dx = \frac{1}{2}(-3) = -\frac{3}{2}$ . Therefore, $\int_{8}^{5} \frac{1}{2}s(x)\,dx = -(-\frac{3}{2}) = \frac{3}{2}$ . A common mistake is applying only one property, giving either $-\frac{3}{2}$ or $-6$ . Always apply properties systematically: factor out constants, then handle limit reversals.

Question 6

At $x=-14$ , $g'(-14)=0$ and $g''(-14)<0$ ; what does the Second Derivative Test imply?

Local minimum at $x=-14$ .
Inflection point at $x=-14$ .
Local maximum at $x=-14$ . (correct answer)
Cannot be determined from the information given.
Neither a maximum nor minimum because $g'(-14)=0$ .

Explanation: The Second Derivative Test determines critical point nature through concavity analysis at those points. Since $g'(-14)=0$ establishes a critical point and $g''(-14)<0$ indicates negative concavity, the function is concave down at $x=-14$ . Concave down behavior at a critical point forms a hill-shaped curve, indicating a local maximum. Choice A could confuse students who think negative signs correspond to minimums, but negative second derivative specifically means downward concavity (maximum). Apply this test when you have verified both the critical point condition and the sign of the second derivative.

Question 7

A particle has acceleration $a(t)=2t$ with $v(0)=0$ and $x(0)=0$ . What is the displacement on $[0,2]$ ?

2
8
4
$\dfrac{8}{3}$ (correct answer)
$\dfrac{4}{3}$

Explanation: This problem tests the skill of using integrals to reason about motion, specifically finding displacement from acceleration through double integration. v(t) = ∫0 to t 2s ds = t², then displacement = ∫0 to 2 s² ds = [s³/3] from 0 to 2 = 8/3. This connects acceleration to velocity to position. The process accumulates changes hierarchically. A tempting distractor like 4 might come from integrating acceleration directly without finding velocity first, but that skips a step. A transferable strategy is to interpret the definite integral of velocity as the net area under the curve, representing displacement.

Question 8

A function is $d(x)=\sin\!\left(\sqrt{3x+7}\right)$ . What is $d'(x)$ ?

$\cos\!\left(\sqrt{3x+7}\right)$
$\dfrac{3}{2\sqrt{3x+7}}\cos\!\left(\sqrt{3x+7}\right)$ (correct answer)
$\dfrac{1}{2\sqrt{3x+7}}\cos\!\left(\sqrt{3x+7}\right)$
$\dfrac{3}{2\sqrt{3x+7}}\sin\!\left(\sqrt{3x+7}\right)$
$\dfrac{3}{\sqrt{3x+7}}\cos\!\left(\sqrt{3x+7}\right)$

Explanation: Sine of square root needs chain rule. Outer sin(u), u = √(3x + 7), derivative cos(u), inner u' = (1/2)(3x + 7)^{-1/2} * 3 = 3/(2 √(3x + 7)). d'(x) = cos(√(3x + 7)) * 3/(2 √(3x + 7)). Common omission: missing 1/2 or 3. Incorrect exponent handling. Spot trig of radical linear and layer chain. This transfers to root composites.

Question 9

A differentiable function $S(t)$ models savings. Which expression represents the instantaneous rate savings changes at $t=0$ ?

$\dfrac{S(1)-S(0)}{1}$
$S(0)$
$S'(0)$ (correct answer)
$\dfrac{S(0)-S(-1)}{1}$
$\dfrac{S(t)}{t}$ at $t=0$

Explanation: This question involves interpreting derivative notation in the context of savings rate change. When S(t) models savings as a function of time, the instantaneous rate at which savings changes at t=0 is the derivative S'(0). This notation represents how savings changes instantaneously with respect to time at that specific moment. In financial contexts, this represents the instantaneous savings rate or the rate of accumulation/depletion at t=0. Choice A represents the average rate of change of savings between t=0 and t=1, which doesn't capture the instantaneous behavior specifically at t=0. For financial rates and instantaneous changes, use derivative notation to capture precise rates at specific times.

Question 10

What is the correct volume integral setup if the base is bounded by $y=\sin x$ and $y=0$ on $[0,\pi]$ with square cross sections perpendicular to the $x$ -axis?

$\displaystyle \int_{0}^{\pi} \sin x\,dx$
$\displaystyle \int_{0}^{\pi} (\sin x)^2\,dx$ (correct answer)
$\displaystyle \int_{0}^{1} (\arcsin y)^2\,dy$
$\displaystyle \int_{0}^{\pi} (\pi-\sin x)^2\,dx$
$\displaystyle \int_{0}^{\pi} (\cos x)^2\,dx$

Explanation: This problem requires using the method of cross-sectional volumes to find the volume of a solid with square cross sections. The base of the solid is the region bounded by y = sin x and y = 0 on [0, π]. For cross sections perpendicular to the x-axis, the side length of each square is the vertical distance between the curves, which is sin x. Therefore, the area of each cross section is (sin x)². Integrating this area function from 0 to π gives the volume as ∫ from 0 to π of (sin x)² dx. A tempting distractor is choice A, which forgets to square the side length and computes the area under the sine curve instead. In general, when modeling volumes with known cross sections, identify the axis of integration, determine the varying dimension of the cross section as a function of the integration variable, compute the area function, and integrate over the appropriate limits.

Question 11

A diffusion model uses $\int \frac{1}{x^2+10x+29}\,dx$ . Which is an antiderivative?

$\frac{1}{2}\arctan\!\left(\frac{x+5}{2}\right)+C$ (correct answer)
$2\arctan\!\left(\frac{x+5}{2}\right)+C$
$\arctan\!\left(\frac{x+5}{2}\right)+C$
$\ln(x^2+10x+29)+C$
$\frac{1}{2}\ln|x^2+10x+29|+C$

Explanation: The key skill here is completing the square as preparatory algebra for integrating rational functions with quadratic denominators. For the integrand 1/(x² + 10x + 29), the quadratic does not factor over the reals since the discriminant is negative. Completing the square rewrites it as (x + 5)² + 4, which matches the form u² + a² with a = 2. This enables the use of the integral formula ∫ du/(u² + a²) = (1/a) arctan(u/a) + C, resulting in (1/2) arctan((x + 5)/2) + C. A tempting distractor is ln(x² + 10x + 29) + C, but it fails because logarithmic antiderivatives apply to factorable quadratics, not this irreducible case. Always compute the discriminant of the quadratic denominator first; if negative, complete the square and apply the arctangent formula for accurate integration.

Question 12

Let $y$ be defined by $e^{y}+xy=4$ . Which differentiation procedure is needed to find $\dfrac{dy}{dx}$ ?

Quotient rule
Implicit differentiation (including product rule on $xy$ ) (correct answer)
Power rule only
Chain rule only on an explicit function
Logarithmic differentiation

Explanation: This problem involves selecting the differentiation procedure for finding $\dfrac{dy}{dx}$ when $y$ is defined implicitly by $e^{y}+xy=4$ . Since $y$ cannot be easily isolated as an explicit function of $x$ , implicit differentiation is required. This involves differentiating both sides with respect to $x$ , treating $y$ as a function of $x$ . The term $e^y$ requires the chain rule (giving $e^y \frac{dy}{dx}$ ), and the term $xy$ requires the product rule (giving $y + x\frac{dy}{dx}$ ). Neither quotient rule alone nor explicit function methods apply to this implicit relationship. When you have an equation mixing $x$ and $y$ terms where $y$ cannot be isolated, implicit differentiation with appropriate subsidiary rules is necessary.

Question 13

Let $s$ be continuous on $[4,8]$ with $s(4)=11$ and $s(8)=15$ . What must occur?

There exists $c\in(4,8)$ such that $s(c)=14$ . (correct answer)
There exists $c\in(4,8)$ such that $s(c)=16$ .
There exists $c\in(8,10)$ such that $s(c)=14$ .
There exists $c\in(4,8)$ such that $s(c)=10$ .
A solution is not guaranteed because $s(4)$ and $s(8)$ are both positive.

Explanation: The Intermediate Value Theorem (IVT) for continuous s on [4, 8] with s(4) = 11 and s(8) = 15 guarantees values between 11 and 15. Since 11 < 14 < 15, there exists c in [4, 8] with s(c) = 14, and as 14 ≠ 11, 14 ≠ 15, c is in (4, 8). This applies even with same-sign positive endpoints. A common error is thinking same signs prevent guarantees, but IVT focuses on the range. Selecting 16 or 10, outside the range, is incorrect. Transferable IVT checklist: 1. Confirm the function is continuous on [a, b]. 2. Verify k is between f(a) and f(b). 3. If yes, there exists c in [a, b] with f(c) = k; if k is strictly between and ≠ f(a), ≠ f(b), then c is in (a, b).

Question 14

An antiderivative of $u(x)$ is $U(x)=\cos x-2x$ . Compute $\int_{\pi/2}^{\pi} u(x)\,dx$ .

$U(\pi/2)-U(\pi)$
$U(\pi)$
$U(\pi)+U(\pi/2)$
$U(\pi)-U(\pi/2)$ (correct answer)
$U(\pi/2)$

Explanation: This question requires applying the Fundamental Theorem of Calculus Part 2 to evaluate a definite integral when given an antiderivative. Since U(x) = cos x - 2x is an antiderivative of u(x), we find ∫[π/2 to π] u(x)dx by computing U(π) - U(π/2). Evaluating: U(π) = cos(π) - 2π = -1 - 2π and U(π/2) = cos(π/2) - 2(π/2) = 0 - π = -π, so U(π) - U(π/2) = -1 - 2π - (-π) = -1 - π. Option A, U(π/2) - U(π), would give 1 + π, which has the wrong sign due to incorrect ordering. Always apply FTC Part 2 as F(upper) - F(lower) to maintain the correct orientation.

Question 15

Let $A(t)$ be the area of a growing stain. Which quantity is the instantaneous rate of area change at $t=5$ ?

$A(6)-A(5)$
$\dfrac{A(7)-A(3)}{7-3}$
$\dfrac{A(5)-A(0)}{5}$
$\lim_{h\to 0}\dfrac{A(5+h)-A(5)}{h}$ (correct answer)
$A(5)-A(4)$

Explanation: The instantaneous rate of area change at $t=5$ is the derivative of area with respect to time, measuring exactly how fast the stain is growing at that moment. Choice D correctly represents this with $\lim_{h\to 0}\frac{A(5+h)-A(5)}{h}$ , which is the limit definition of the derivative at $t=5$ . Choices B and C calculate average rates of growth over various intervals, while choices A and E compute area differences without dividing by time, giving square units rather than square units per time. A common mistake is thinking that the change over one time unit, like $A(5)-A(4)$ , approximates the instantaneous rate, but this is still an average over that interval. Remember: instantaneous rates require limits to capture the exact rate at a single instant, while any calculation over a finite interval yields an average rate.

Question 16

The relation $x\sqrt{y}+y\sqrt{x}=10$ defines $y$ implicitly. What is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?

$\dfrac{dy}{dx}=-\dfrac{\sqrt{y}+\frac{y}{2\sqrt{x}}}{\frac{x}{2\sqrt{y}}+\sqrt{x}}$ (correct answer)
$\dfrac{dy}{dx}=-\dfrac{\sqrt{y}+\frac{y}{2\sqrt{x}}}{\frac{x}{2\sqrt{y}}}$
$\dfrac{dy}{dx}=-\dfrac{\sqrt{y}}{\sqrt{x}}$
$\dfrac{dy}{dx}=-\dfrac{\frac{x}{2\sqrt{y}}+\sqrt{x}}{\sqrt{y}+\frac{y}{2\sqrt{x}}}$
$\dfrac{dy}{dx}=-\dfrac{\sqrt{y}+\frac{1}{2\sqrt{x}}}{\frac{x}{2\sqrt{y}}+\sqrt{x}}$

Explanation: This problem requires implicit differentiation to find $\dfrac{dy}{dx}$ for the relation $x \sqrt{y} + y \sqrt{x} = 10$ . Differentiating both sides gives $\sqrt{y} + x \left( \frac{1}{2 \sqrt{y}} \right) \dfrac{dy}{dx} + \sqrt{x} \dfrac{dy}{dx} + y \left( \frac{1}{2 \sqrt{x}} \right) = 0$ , using product and chain rules for each term. The $\dfrac{dy}{dx}$ terms are $\left( \frac{x}{2 \sqrt{y}} \right) \dfrac{dy}{dx} + \sqrt{x} \dfrac{dy}{dx}$ , grouped, while $- \sqrt{y} - \frac{y}{2 \sqrt{x}}$ is on the other side. Solving yields $- \frac{ \sqrt{y} + \frac{y}{2 \sqrt{x}} }{ \frac{x}{2 \sqrt{y}} + \sqrt{x} }$ , matching the form. Choice B is tempting if someone forgets the $\sqrt{x} \dfrac{dy}{dx}$ term, omitting it in the denominator. To recognize implicit differentiation, look for roots or other functions mixing $x$ and $y$ implicitly.

Question 17

Let $m(x)=\sqrt{\sec(2x)}$ . What is $m'(x)$ ?

$\dfrac{\sec(2x)\tan(2x)}{\sqrt{\sec(2x)}}$
$\dfrac{\sec(2x)\tan(2x)}{\sqrt{\sec(2x)}}\cdot 2$
$\dfrac{\sec(2x)\tan(2x)}{2\sqrt{\sec(2x)}}\cdot 2$ (correct answer)
$\dfrac{\tan(2x)}{2\sqrt{\sec(2x)}}$
$\dfrac{\sec(2x)}{2\sqrt{\sec(2x)}}$

Explanation: m(x) = √(sec(2x)) = [sec(2x)]^{1/2}. Chain: (1/2) [sec(2x)]^{-1/2} · sec(2x) tan(2x) · 2 = sec(2x) tan(2x) / √(sec(2x)). Multiple chains for sec and 2x. Omission: Forgetting the 2. Some miss tan. Pattern: Roots of trig with multiples layer chains.

Question 18

A function is $j(x)=\ln\!\left(\dfrac{1}{x^2+2}\right)$ . What is $j'(x)$ ?

$\dfrac{1}{\frac{1}{x^2+2}}$
$\dfrac{-2x}{x^2+2}$ (correct answer)
$\dfrac{2x}{x^2+2}$
$\dfrac{-2x}{(x^2+2)^2}$
$\ln\!\left(\dfrac{1}{x^2+2}\right)\cdot \left(\dfrac{-2x}{(x^2+2)^2}\right)$

Explanation: Log of reciprocal polynomial uses properties. j(x) = -ln(x^2 + 2), j'(x) = - (2x)/(x^2 + 2). Outer ln, inner 1/(x^2 + 2), but property simplifies. Common omission: sign error from reciprocal. Missing 2x. Identify log of inverse and use negative log then differentiate. This pattern aids in denominator logs.

Question 19

The number of books checked out is $B(t)$ after $t$ hours. Interpret $B'(2)=35$ .

At $t=2$ , 35 books are checked out.
At $t=2$ hours, books checked out are increasing at 35 books per hour. (correct answer)
At $t=2$ hours, hours increase at 35 hours per book.
During the first 2 hours, 35 books were checked out total.
At $t=2$ hours, books checked out increase at 35 books per minute.

Explanation: This question requires interpreting library circulation rates in practical contexts. The derivative B'(2) = 35 represents the instantaneous rate of change of books checked out at t = 2 hours. Since B(t) represents books and t is in hours, B'(2) has units of books per hour. The positive value indicates books are being checked out. Therefore, at t = 2 hours, books checked out are increasing at 35 books per hour. Common errors include confusing the derivative with total books checked out (choice A) or using incorrect time units (choice E states per minute). The strategy is to interpret positive derivatives as rates of activity or process completion while preserving the correct temporal units.

Question 20

For $f(x)=\begin{cases}x+2,&x<0\\2,&x=0\\2-x,&x>0\end{cases}$ , does $f'(0)$ exist, and why?

Yes; the function is continuous at $0$ , so $f'(0)$ exists.
No; the one-sided derivatives at $0$ are different, so $f'(0)$ does not exist. (correct answer)
Yes; the derivative exists because $f(0)$ is defined.
No; the function has a jump discontinuity at $0$ .
Yes; different one-sided derivatives imply a vertical tangent.

Explanation: This question assesses the connection between differentiability and continuity in AP Calculus AB, analyzing piecewise linear functions. For this f(x), it is continuous at x=0 with value 2 matching both limits. However, the left-hand derivative is 1, while the right-hand is -1, creating a corner. This difference means f'(0) does not exist. A tempting distractor might say yes because of continuity, but slopes must match. Remember this differentiability checklist: verify continuity at the point, then check that left- and right-hand derivatives exist and are equal.

Question 21

For $D(t)= -5t^4+0t^3+8t^2-7$ , what is $D'(t)$ ?

$-20t^3+16t$ (correct answer)
$-20t^4+16t^2$
$-5t^4+8t^2-7$
$-5t^3+16t$
$-20t^3+16t-7$

Explanation: Using linearity, differentiate D(t) = -5t^4 + 0t^3 + 8t^2 - 7 term by term. Derivative of -5t^4 is -20t^3, of 0t^3 is 0, of 8t^2 is 16t, of -7 is 0. Thus, D'(t) = -20t^3 + 16t. A common misuse is treating zero-coefficient terms as contributing, though they don't. Sign handling in leading terms is critical. For such functions, apply sum rules and power rule systematically via linearity.

Question 22

Let $s(x)=x^3-2$ and $s(2)=6$ . Find $(s^{-1})'(6)$ .

$12$
$\dfrac{1}{12}$ (correct answer)
$\dfrac{1}{s'(6)}$
$s'(2)$
$\dfrac{1}{6}$

Explanation: This problem tests the skill of differentiating inverse functions. The derivative of the inverse function at a point a is given by 1 over the derivative of the original function at b, where s(b) = a. Here, s(2) = 6, so b = 2 and a = 6. Compute s'(x) = 3x², so s'(2) = 12, thus (s⁻¹)'(6) = 1/12. A tempting distractor is A, 12, which might be chosen if one forgets to take the reciprocal, but that gives the slope of the tangent to s, not to s⁻¹. To solve similar problems, always remember that the derivatives of a function and its inverse are reciprocals at corresponding points.

Question 23

Let $f$ be continuous on $[-1,1]$ with $f(-1)=2$ and $f(1)=2$ ; which statement must be true?

$f$ has an absolute maximum at both endpoints.
$f$ has an absolute minimum at $x=0$ .
$f$ attains an absolute maximum and an absolute minimum on $[-1,1]$ . (correct answer)
$f$ must have exactly one critical point.
Equal endpoint values prevent existence of absolute extrema.

Explanation: This problem applies the Extreme Value Theorem to determine what must be guaranteed for continuous functions on closed intervals with specific endpoint conditions. Since $f$ is continuous on the closed interval $[-1,1]$ , the EVT unconditionally guarantees that $f$ must attain both an absolute maximum value and an absolute minimum value somewhere on this interval. The condition that $f(-1)=f(1)=2$ provides information about the endpoint values but doesn't change the fundamental EVT guarantee or prevent the existence of absolute extrema. Equal endpoint values simply mean both endpoints have the same function value - they could both be absolute extrema, neither could be, or one could be depending on the function's interior behavior. Choice E incorrectly suggests equal endpoint values prevent absolute extrema. The core EVT truth: continuous function on closed interval always has absolute maximum and minimum values.

Question 24

If $A(5)=0$ and $A'(5)=2$ , use local linearity at $5$ to approximate $A(5.4)$ .

$0.8$ (correct answer)
$2.4$
$0.4$
$0.08$
$-0.8$

Explanation: This problem tests linearization near x = 5. For A(5) = 0 and A'(5) = 2, A(5.4) ≈ 0 + 2(0.4) = 0.8. Conceptually, it's building from zero with positive slope. Common mistake: Using Δx = 4. Mistake: Subtracting instead. Strategy: Start from f(a) and add the scaled derivative.

Question 25

A model for distance is $d(t)=2t^8-3t^2+5t^4$ . What is $d'(t)$ ?

$16t^7-6t+20t^3$ (correct answer)
$16t^8-6t^2+20t^4$
$2t^7-3t+5t^3$
$16t^7-3t+20t^3$
$16t^7-6t+5t^4$

Explanation: This problem requires applying the power rule to a polynomial with terms out of order. The power rule states that d/dt[t^n] = nt^(n-1). For d(t) = 2t^8 - 3t^2 + 5t^4, we differentiate each term independently: the derivative of 2t^8 is 2·8t^7 = 16t^7, the derivative of -3t^2 is -3·2t^1 = -6t, and the derivative of 5t^4 is 5·4t^3 = 20t^3. Therefore, d'(t) = 16t^7 - 6t + 20t^3. Choice E incorrectly differentiates the last term as 5t^4 instead of 20t^3, keeping the original term unchanged. When differentiating polynomials, treat each term separately and don't be distracted by their order in the expression.

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