Solving Nonlinear Functions - PSAT Math

What is the range of the function $y=\sqrt{x-9}$ in real numbers? Choose the set of all possible output values.
Answer: $[0,\infty)$ The question asks for the range of y = √(x - 9) over real numbers. This is a square root function, nonlinear and increasing from a minimum. Domain x ≥ 9 gives y ≥ √0 = 0, and as x → ∞, y → ∞, so range [0, ∞). This includes 0 and all positive reals. Matches choice B; [9, ∞) is the domain, not range. Key error: Confusing domain with range. Strategy: Solve for possible y by considering the inverse or output limits for square root functions.
Which of the following lists all real solutions to the equation $x^{2}-7x+12=0$ ?
Answer: $x=3 \text{ or } x=4$ When you encounter a quadratic equation like $x^2 - 7x + 12 = 0$ , you have several solution methods available: factoring, completing the square, or using the quadratic formula. For this equation, factoring is the most efficient approach. To factor $x^2 - 7x + 12$ , you need two numbers that multiply to 12 (the constant term) and add to -7 (the coefficient of the middle term). Those numbers are -3 and -4, since $(-3) \times (-4) = 12$ and $(-3) + (-4) = -7$ . This gives you $(x - 3)(x - 4) = 0$ . Using the zero product property, if $(x - 3)(x - 4) = 0$ , then either $x - 3 = 0$ or $x - 4 = 0$ . Solving these gives $x = 3$ or $x = 4$ , which matches choice D. Choice A represents what you'd get from incorrectly applying the quadratic formula with $b^2 - 4ac = 49 - 48 = 1$ , but making an error in simplification. Choice B gives the correct numbers but with wrong signs—this happens when students confuse the factors $(-3)$ and $(-4)$ with the actual solutions. Choice C mixes one correct solution ( $x = 4$ ) with one incorrect sign ( $x = -3$ instead of $x = 3$ ). Always verify your solutions by substituting back into the original equation. For $x = 3$ : $9 - 21 + 12 = 0$ ✓. For $x = 4$ : $16 - 28 + 12 = 0$ ✓. Strategy tip: When factoring quadratics, double-check that your factors multiply to give the constant term and add to give the middle coefficient—this catches most factoring errors before you solve.
Let $f(x)=\begin{cases}x^{2}-4 & \text{if }x\ge0\\2x+1 & \text{if }x<0\end{cases}.$ For what value of $x$ does $f(x)=5$ ?
Answer: $x=3$ Piecewise functions require you to determine which piece of the function applies based on the given condition, then solve within that piece. Since we need $f(x) = 5$ , we must check both pieces of this function. For $x \geq 0$ , we use $f(x) = x^2 - 4$ . Setting this equal to 5: $x^2 - 4 = 5$ , so $x^2 = 9$ , which gives us $x = \pm 3$ . However, since this piece only applies when $x \geq 0$ , we can only use $x = 3$ . For $x < 0$ , we use $f(x) = 2x + 1$ . Setting this equal to 5: $2x + 1 = 5$ , so $2x = 4$ , which gives us $x = 2$ . But since $x = 2$ is positive, it doesn't satisfy the condition $x < 0$ for this piece of the function. Therefore, only $x = 3$ works, making C the correct answer. Choice A ( $x = \pm 3$ ) includes both solutions from solving $x^2 - 4 = 5$ , but ignores the domain restriction that $x \geq 0$ for that piece. Choice B ( $x = -2$ ) appears to come from solving $2x + 1 = 5$ incorrectly or confusing the setup. Choice D (no real solution) would be wrong since we do have $x = 3$ as a valid solution. Strategy tip: Always check that your solution satisfies the domain condition for the piece of the piecewise function you're using. A mathematically correct solution that violates the domain restriction isn't valid for that piece.
If $f(x)=\sqrt{16x}$ , what is $f(\frac{1}{4})$ ? Simplify the square root exactly.
Answer: $2$ This question asks for the value of f(1/4) where f(x) = $\sqrt{16x}$ , requiring exact simplification. The function is a square root function, a type of radical nonlinear function, defined for $x \geq 0$ and producing non-negative outputs. Substitute x=1/4: $\sqrt{16 \times \frac{1}{4}} = \sqrt{4} = 2$ , simplifying by factoring 16 as 4*4 and taking $\sqrt{4} = 2$ . Ensure the expression under the root is non-negative, which it is here. A common error is miscalculating $\sqrt{16/4} = \sqrt{4} = 2$ but forgetting to simplify properly, or thinking it's $\sqrt{16} \times \sqrt{\frac{1}{4}} = 4 \times \frac{1}{2} = 2$ , which confirms. When simplifying radicals, factor out perfect squares for exact values.
A function satisfies $f(0)=7$ and for each increase of 1 in $x$ , the output is multiplied by 4. Which equation defines $f(x)$ ? Choose the exponential function with initial value 7.
Answer: $f(x)=7\cdot 4^x$ This question asks for the equation defining f(x) given f(0)=7 and multiplication by 4 for each unit increase in x. The function is exponential growth, as the output multiplies by a constant factor, unlike linear addition or polynomial powers. Starting with f(0)=7, then f(1)=7*4, f(2)=7* $4^2$ , so generally f(x)=7* $4^x$ . Choice B has the base and initial swapped. A common error is selecting a linear model like 7+4x, which adds instead of multiplies. For recursive descriptions, convert to explicit exponential form by identifying the initial value and growth factor.
A bacteria culture starts with 500 cells and doubles every 3 hours. The number of cells after $t$ hours is modeled by $N(t)=500\cdot 2^{t/3}$ . Which expression gives the number of cells after 9 hours, and what is its value?
Answer: $500\cdot 2^{9/3}=4000$ This question asks us to find the number of bacteria cells after 9 hours using the given exponential growth model. The function N(t) = 500· $2^{t/3}$ is an exponential function where the bacteria population doubles every 3 hours. To find the number of cells after 9 hours, we substitute t = 9 into the formula: N(9) = 500· $2^{9/3}$ = 500· $2^3$ = 500·8 = 4000. A common error is to use $2^9$ instead of $2^{9/3}$ , which would give an incorrect answer of 256,000 cells. When dealing with exponential growth problems, always check that the exponent correctly reflects the given time period and growth rate.
A bacteria culture starts with 200 cells and doubles every 3 hours. The number of cells after $t$ hours is modeled by a nonlinear function. Which equation represents the number of cells after $t$ hours? Be careful to use multiplicative growth rather than adding a constant amount each hour.
Answer: $N(t)=200\cdot 2^{t/3}$ The question asks for the equation modeling the number of bacteria cells after t hours, starting with 200 cells and doubling every 3 hours. This is an exponential growth function, a type of nonlinear function where the quantity multiplies by a constant factor over time. To find the model, recognize that doubling every 3 hours means the growth factor is 2 raised to the power of t/3, so N(t) = 200 * $2^{t/3}$ . Verify by checking at t=3: N(3) = 200 * $2^1$ = 400, which is double the initial amount. A key error is choosing linear growth like choice A, which adds a constant instead of multiplying, failing to capture exponential doubling. Another mistake is swapping the base and initial value as in choice C, which grows too rapidly. For growth problems, distinguish exponential from linear by checking if changes are multiplicative rather than additive.
A store's online sales follow $S(t)=150\cdot 1.2^t$ , where $t$ is weeks. Which best describes the growth rate? Interpret $1.2$ as a weekly multiplier.
Answer: 20% increase/week This question asks for the best description of the growth rate in the given exponential sales model. The function is an exponential growth function since the base 1.2 is greater than 1, distinguishing it from decay functions where the base is between 0 and 1. To find the growth rate, interpret the base 1.2 as a multiplier, meaning sales increase by a factor of 1.2 each week, which corresponds to a 20% increase because 1.2 = 1 + 0.2. Subtracting 1 from the base and converting to a percentage gives the rate directly. A common error is confusing the multiplier with the percentage increase, such as thinking 1.2 means 120% increase, which would actually be a multiplier of 2.2. When dealing with exponential models, always express the growth rate as (base - 1) × 100% for percentage interpretation.
A coordinate plane shows a function that is always positive, passes through $(0,2)$ , and has horizontal asymptote $y=0$ . It also passes through $(-1,4)$ . Which equation matches?
Answer: $y=2\left(\tfrac12\right)^x$ The question requires the equation for a positive function through (0,2) with y=0 asymptote and through (-1,4). This is an exponential decay function, nonlinear approaching zero asymptotically. Points give: at x=0, a=2; at x=-1, 2*(1/b)=4, b=1/2, so y=2* $(1/2)^x$ . Matches points and asymptote. Choice A fits; reciprocal lacks exponential shape. Common error: Choosing growth for increasing at negative x. Strategy: Use points from both sides of y-intercept to find base in exponentials.
For the rational function $r(x)=\dfrac{2x-1}{x-4}$ , what is $r(6)$ ? Substitute carefully and simplify the fraction.
Answer: $\tfrac{11}{2}$ The question asks for r(6) where $r(x) = \frac{2x - 1}{x - 4}$ . This is a rational function, nonlinear and evaluated by substitution. Plug in x=6: $\frac{12 - 1}{6 - 4} = \frac{11}{2}$ . Simplify the fraction after substituting. A common error is mis-substituting like getting $\frac{1}{11}$ . Another is confusing numerator and denominator. For rationals, substitute carefully and simplify to avoid arithmetic errors.
A coordinate plane shows a curve with a starting point at $(-4,0)$ and it increases to the right, passing through $(0,2)$ . The graph exists only for $x\ge -4$ . Which equation matches?
Answer: $y=\sqrt{x+4}$ The question requires the equation for a graph starting at (-4,0), increasing rightward through (0,2), defined for x ≥ -4. This is a square root function, nonlinear with a gradually decreasing rate of increase. The domain x ≥ -4 suggests √(x + 4), as the argument must be nonnegative. Testing points: at x=-4, √0=0; at x=0, √4=2, matching the graph. This is choice A; others like √(4 - x) would have domain x ≤ 4, not fitting. Common errors include mixing up the shift direction in the argument. Strategy: Use domain restrictions and key points to distinguish square root from absolute value or quadratic functions.
A coordinate plane shows an exponential curve that passes through $(0,5)$ and increases, passing through $(1,15)$ . Which equation matches this graph?
Answer: $y=5\cdot 3^x$ The question seeks the equation for an exponential curve passing through (0,5) and (1,15), increasing. This is an exponential growth function, nonlinear with a constant growth factor. At x=0, y=5 (initial); at x=1, 5 b =15, b=3, so y=5 $3^x$ . Testing: matches points, increases. Choice B fits; linear like D passes points but is not exponential. Common error: Choosing linear for two points without noting exponential requirement. Strategy: Determine growth factor from points to build exponential equations, distinguishing from linear trends.
Let $f(x)=|2x-5|$ . What is the value of $f(1)$ ?
Answer: $3$ This question asks us to evaluate an absolute value function at a specific point. To find f(1) where f(x) = |2x-5|, we substitute x = 1: f(1) = |2(1)-5| = |2-5| = |-3| = 3. The absolute value operation takes the distance from zero, so |-3| = 3. A common error is to forget to apply the absolute value after evaluating the expression inside, which would give -3 instead of 3. Always complete all operations inside the absolute value bars first, then take the absolute value of the result.
A coordinate plane shows a square-root curve that begins at $(0,-3)$ and increases to the right, passing through $(4,-1)$ . Which equation matches the graph?
Answer: $y=\sqrt{x}-3$ The question seeks the equation for a square-root curve starting at $(0,-3)$ through $(4,-1)$ , increasing rightward. This is a square root function, nonlinear with concave down shape. Form $y = \sqrt{x} - 3$ : at $x=0$ , $y=-3$ ; at $x=4$ , $\sqrt{4} - 3 = 2 - 3 = -1$ , matches. Domain $x \geq 0$ fits starting point. Choice A; absolute value is V-shaped. Common error: Adding instead of subtracting the shift. Strategy: Use starting point and another to test shifts in square root equations.
Which is the domain of $f(x)=\sqrt{4x+1}$ ? Give the set of all real $x$ values for which the function is defined.
Answer: $\left[-\tfrac14,\infty\right)$ The question requires the domain of f(x) = √(4x + 1). This is a square root function, nonlinear and defined for nonnegative arguments. Solve 4x + 1 ≥ 0: 4x ≥ -1, x ≥ -1/4. Includes x = -1/4 where f=0. Matches choice B; open interval excludes the endpoint. Common error: Forgetting to include the equality for zero. Strategy: Solve inequalities carefully for square roots, noting closed intervals when zero is allowed.
For the absolute value function $v(x)=|x|-4$ , which is the correct range? Consider the smallest possible output value and whether outputs can increase without bound.
Answer: $[-4,\infty)$ This question asks for the correct range of the absolute value function v(x)=|x|-4. The function is absolute value, producing a V-shape with minimum at x=0, distinguishing from radicals or exponentials. The output |x| ≥ 0, so |x|-4 ≥ -4, achieving -4 at x=0, and as |x| increases, v(x) → ∞. Thus, the range is all values from -4 inclusive to infinity. A key error is excluding the minimum like in choice C, or reversing the interval. When finding range, identify the vertex minimum and note unbounded increase.
For the radical function $h(x)=\sqrt{2x-5}$ , what is the domain in real numbers? Choose the interval of all $x$ values that make the expression under the radical nonnegative.
Answer: $[\tfrac{5}{2},\infty)$ The question asks for the domain of the radical function $h(x) = \sqrt{2x-5}$ in real numbers. This is a square root function, nonlinear and defined only where the radicand is nonnegative. Solve $2x - 5 \geq 0$ to get $x \geq \frac{5}{2}$ , including the endpoint where $h\left(\frac{5}{2}\right) = 0$ . The interval is $\left[\frac{5}{2}, \infty\right)$ . A common error is excluding the endpoint as in C, but square root of zero is defined. Another mistake is reversing the inequality like D. For radicals, always solve the inequality to find the domain and distinguish from unrestricted functions like exponentials.
If $P(t)=1200(0.9)^t$ models a population after $t$ years, what is the percent change each year? Interpret the base of the exponential function as a growth or decay factor.
Answer: 10% decrease The question asks for the annual percent change in the population model P(t) = 1200 $(0.9)^t$ . This is an exponential decay function, nonlinear with multiplicative changes over time. The base 0.9 means each year the population is 90% of the previous, a 10% decrease. Interpreting 0.9 as 1 - 0.1 confirms the decay rate. This matches choice A; options like 90% decrease misread the factor. A key error is confusing the base with a growth percentage. Strategy: For exponential models, subtract the base from 1 and multiply by 100 to find percent change, noting if it's growth or decay.
A coordinate plane shows a curve that is symmetric about the y-axis, passes through $(0,0)$ , and has points $(2,4)$ and $(-2,4)$ . Which equation matches the graph?
Answer: $y=x^2$ The question requires the equation for a graph symmetric about the y-axis, passing through (0,0), (2,4), and (-2,4). This is a quadratic function, nonlinear and even, like y = x². Symmetry about y-axis means f(-x) = f(x); testing y = x²: at x=0, 0; x=2, 4; x=-2, 4, matches. Absolute value y=|x| passes (0,0) but (2,2) not 4. This is choice A. Common error: Selecting absolute value without verifying points. Strategy: Check symmetry and plug in points to distinguish even quadratics from other symmetric nonlinear functions like absolute value.
The polynomial $f(x)=x^{3}-4x^{2}-7x+10$ is known to have a root at $x=5$ . Which of the following lists all real solutions of $f(x)=0$ ?
Answer: $x=5,\;x=1,\;x=-2$ When you encounter a polynomial with a known root, you can use polynomial division or factoring to find all remaining roots. Since we know $x = 5$ is a root of $f(x) = x^3 - 4x^2 - 7x + 10$ , we know that $(x - 5)$ is a factor. To find the complete factorization, divide the polynomial by $(x - 5)$ . Using polynomial long division or synthetic division, we get: $f(x) = (x - 5)(x^2 + x - 2)$ Now we need to factor the quadratic $x^2 + x - 2$ . We're looking for two numbers that multiply to $-2$ and add to $1$ . Those numbers are $2$ and $-1$ , so: $x^2 + x - 2 = (x + 2)(x - 1)$ Therefore: $f(x) = (x - 5)(x + 2)(x - 1)$ Setting each factor equal to zero gives us all three roots: $x = 5$ , $x = -2$ , and $x = 1$ . Looking at the answer choices: Choice A gives $x = 5, 1, 2$ (missing the negative sign on the third root). Choice B gives $x = 5, -1, 2$ (wrong signs on both remaining roots). Choice C gives $x = 5, -1, -2$ (wrong sign on the second root). Choice D correctly lists $x = 5, 1, -2$ . Study tip: When given one root of a polynomial, always factor it out completely to find all roots. Double-check your factorization by expanding it back to verify you get the original polynomial. Pay careful attention to positive and negative signs in your final answer.

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