AP Calculus AB

AP Calculus AB Practice Test: Practice Test 9

Practice Test 9 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

Velocity is positive and decreasing at $t=5$ ; which statement about acceleration at $t=5$ is correct?

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Question 1

Velocity is positive and decreasing at $t=5$ ; which statement about acceleration at $t=5$ is correct?

Acceleration is negative (correct answer)
Acceleration is positive
Acceleration is zero
Acceleration equals velocity
Acceleration cannot be determined

Explanation: This straight-line motion problem connects velocity behavior to acceleration using derivative relationships. When velocity is positive and decreasing at $t = 5$ , we have $v(5) > 0$ and $v'(5) < 0$ . Since acceleration is the derivative of velocity, $a(5) = v'(5) < 0$ , so acceleration is negative. Students might think positive velocity means positive acceleration, but decreasing velocity always means negative acceleration. The key motion analysis strategy is that acceleration equals the derivative of velocity, so decreasing velocity always indicates negative acceleration.

Question 2

For a curve $y=f(x)$ , what does $\displaystyle \lim_{h\to 0}\frac{f(2+h)-f(2)}{h}$ represent?

Average rate of change on $[0,2]$ .
Slope of the tangent line at $x=2$ . (correct answer)
Slope of the secant line from $x=2$ to $x=2+h$ for a fixed $h$ .
The y-intercept of the graph.
The value $f(2+h)-f(2)$ when $h=1$ .

Explanation: The limit given is the derivative $f'(2)$ , which is the slope of the tangent at $x=2$ , choice B, representing instantaneous rate, unlike average which is secant slope over an interval. Choice A is average over $[0,2]$ , and choice C is secant for fixed h. A common confusion is mistaking net change (choice E) or y-intercept (choice D) for the rate. The limit makes it instantaneous. For comparison, link limits to tangents for instantaneous and fixed intervals to secants in curve analyses.

Question 3

A function satisfies $\dfrac{dy}{dx}=\dfrac{y}{x}$ . What is the general solution for $x\ne 0$ ?

$\ln|y|=\dfrac{1}{x}+C$
$y=Cx$ (correct answer)
$y=x+C$
$y=Ce^{x}$
$y=\ln|x|+C$

Explanation: This problem requires solving a differential equation using separation of variables. For dy/dx = y/x with x ≠ 0, separate as dy/y = dx/x, assuming y ≠ 0. Integrate: ln|y| = ln|x| + C. Simplify to y = C x. The distractor y = x + C fails by suggesting addition instead of the proportional solution from logs. To recognize separable equations, look for the form dy/dx = f(x) g(y), which allows rewriting as (1/g(y)) dy = f(x) dx for integration.

Question 4

Let $J(t)= -t^6+16t^2-3t+4$ . What is $J'(t)$ ?

$-6t^5+32t-3$ (correct answer)
$-6t^6+32t^2-3t$
$-t^6+16t^2-3t+4$
$-t^5+32t-3$
$-6t^5+32t+4$

Explanation: Linearity rules enable us to differentiate sums and differences term by term, pulling out constants. For J(t) = -t^6 + 16t^2 - 3t + 4, differentiate each part: derivative of -t^6 is -6t^5, of 16t^2 is 32t, of -3t is -3, and of +4 is 0. Thus, J'(t) = -6t^5 + 32t - 3. A common rule misuse is neglecting to reduce the exponent by 1 while multiplying by the original exponent, resulting in incorrect powers or coefficients. Signs must be carefully preserved during differentiation. To select rules effectively, recognize polynomials and use the sum/difference rule combined with the constant multiple and power rules term by term.

Question 5

Which integral gives the volume when the region under $y=x^2$ from $x=0$ to $x=2$ is revolved about the $x$ -axis?

$\pi\displaystyle\int_{0}^{2}x^{2}\,dx$
$\pi\displaystyle\int_{0}^{2}(x^2)^{2}\,dx$ (correct answer)
$\pi\displaystyle\int_{0}^{2}(2-x^2)^{2}\,dx$
$\pi\displaystyle\int_{0}^{2}(x^2)^{2}\,dy$
$2\pi\displaystyle\int_{0}^{2}x^2\,dx$

Explanation: This problem uses the disc method for revolving around the x-axis. When rotating the region under y = x² from x = 0 to x = 2 around the x-axis, we form circular discs perpendicular to the x-axis. Each disc has radius equal to the y-value at that x-position, which is x², so the disc's area is π(x²)². The volume is found by integrating these disc areas: π∫₀²(x²)²dx = π∫₀²x⁴dx. Choice A incorrectly uses π∫₀²x²dx, forgetting that we need to square the radius function when finding the area of each disc. The key principle: disc method around the x-axis requires π times the square of the function, not just π times the function.

Question 6

A circular bracelet’s radius decreases at $0.05$ cm/s when $r=20$ cm; how fast is its area changing?

$-2\pi\text{ cm}^2/\text{s}$ (correct answer)
$2\pi\text{ cm}^2/\text{s}$
$-0.05\pi\text{ cm}^2/\text{s}$
$-20\pi\text{ cm}^2/\text{s}$
$-4\pi\text{ cm}^2/\text{s}$

Explanation: This problem demonstrates area rate change for a decreasing circular radius. Given A = πr² and dr/dt = -0.05 cm/s (negative for decreasing) when r = 20 cm, taking the derivative: dA/dt = 2πr(dr/dt) = 2π(20)(-0.05) = -2π cm²/s. The negative sign correctly indicates the area is decreasing as the bracelet contracts. Students might forget the negative sign or miscalculate the coefficient.

Question 7

Let $u$ be continuous on $[a,b]$ . If $u$ has no critical points in $(a,b)$ , which statement must be true?

$u$ is constant on $[a,b]$ .
$u$ has no absolute extrema on $[a,b]$ .
Any absolute maximum or minimum of $u$ on $[a,b]$ occurs at $x=a$ or $x=b$ . (correct answer)
$u$ must have at least one local maximum in $(a,b)$ .
$u'(x)$ exists for all $x\in(a,b)$ .

Explanation: This question explores where absolute extrema must occur when there are no interior critical points. If u has no critical points in (a,b), then u has no points where u'(x)=0 or u'(x) doesn't exist in the open interval. By the EVT, u (being continuous on [a,b]) must have absolute extrema, and since these can only occur at critical points or endpoints, they must occur at x=a or x=b. Choice A incorrectly concludes u must be constant, but u could be strictly increasing or decreasing. The extrema location principle: absolute extrema occur at critical points or endpoints; no interior critical points means extrema must be at endpoints.

Question 8

Let $Q(x)=\ln\!\left(1+\sin\!\left(x^2\right)\right)$ . What is $Q'(x)$ ?

$\dfrac{1}{1+\sin(x^2)}$
$\dfrac{2x\cos(x^2)}{1+\sin(x^2)}$ (correct answer)
$\dfrac{\cos(x^2)}{1+\sin(x^2)}$
$2x\cos(x^2)$
$\ln(1+\sin(x^2))\cdot 2x\cos(x^2)$

Explanation: For Q(x) = ln(1 + sin(x²)), chain rule. Outer ln(t), derivative 1/t, t = 1 + sin(x²). t' = cos(x²) * 2x. Thus, Q'(x) = (1/(1 + sin(x²))) * 2x cos(x²). A common omission is missing 2x. Recognize logs of summed trig with inner quadratic and chain accordingly.

Question 9

Define $g(x)=\int_{0}^{x}\dfrac{1}{1+\cos t}\,dt$ where defined. What is $g'(x)$ ?

$\dfrac{1}{1+\cos x}$ (correct answer)
$\dfrac{\sin x}{(1+\cos x)^2}$
$\int_{0}^{x}\dfrac{1}{1+\cos t}\,dt$
$\dfrac{1}{1+\cos t}$
$\left[\int \dfrac{1}{1+\cos t}\,dt\right]_{0}^{x}$

Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function g(x) is defined as the integral from 0 to x of 1/(1+cos t) dt, which accumulates the area under the curve of the integrand from a fixed lower limit to a variable upper limit x. According to FTC Part 1, the derivative g'(x) is simply the integrand evaluated at t = x, so g'(x) = 1/(1+cos x). This holds because differentiating the integral with respect to the upper limit effectively adds the integrand's value at that point, reversing the integration process. A tempting distractor is choice B, which is actually the derivative of choice A and might be selected if someone mistakenly differentiates the result further. To recognize FTC Part 1 applications, look for functions defined as integrals with a variable upper limit and constant lower limit, where the derivative is the integrand at x.

Question 10

A function $g$ has $g'(-2)=0$ and $g''(-2)<0$ . What can be concluded about $x=-2$ ?

Local maximum at $x=-2$ because $g''(-2)<0$ . (correct answer)
Local minimum at $x=-2$ because $g''(-2)<0$ .
Point of inflection at $x=-2$ because $g''(-2)<0$ .
Neither; $g'(-2)=0$ guarantees only an inflection point.
Neither; the Second Derivative Test cannot be used unless $g''(-2)=0$ .

Explanation: This question tests the Second Derivative Test for classifying critical points. We have g'(-2) = 0, establishing a critical point at x = -2. Since g''(-2) < 0, the Second Derivative Test tells us that g has a local maximum at x = -2 because the function is concave down there. The negative second derivative means the graph curves downward like an upside-down U, creating a peak at x = -2. Choice B incorrectly associates negative concavity with a minimum, which is a common sign error. For the AP exam, memorize: f''(c) < 0 means concave down and local max; f''(c) > 0 means concave up and local min.

Question 11

For $r(x)=\dfrac{4}{(x-3)^3}$ , which limit expression matches the behavior of $r(x)$ as $x\to3^-$ ?

$\displaystyle \lim_{x\to3^-} r(x)=-\infty$ (correct answer)
$\displaystyle \lim_{x\to3^-} r(x)=+\infty$
$\displaystyle \lim_{x\to3} r(x)=+\infty$
$\displaystyle r(3)=-\infty$
$\displaystyle \lim_{x\to-3^-} r(x)=-\infty$

Explanation: For r(x) = 4/(x-3)³, we analyze the behavior as x approaches 3 from the left (x→3⁻). When x is slightly less than 3, the denominator (x-3) is a small negative number. Since we're cubing this negative value, (x-3)³ remains negative. The fraction becomes 4/(small negative) = large negative value. Therefore, lim[x→3⁻] r(x) = -∞ correctly represents this behavior. A common error is forgetting that odd powers preserve the sign of negative numbers, unlike even powers. Another mistake is writing r(3) = -∞, which incorrectly suggests the function has a value at x = 3. Notation checklist: Pay attention to the power of the denominator, verify signs carefully for odd vs even powers, and use proper limit notation.

Question 12

For $y=x^3-x$ and $y=0$ on $[-1,1]$ , which setup gives the total area between the curves?

$\displaystyle \int_{-1}^{1}(x^3-x)\,dx$
$\displaystyle \int_{-1}^{0}(x^3-x)\,dx+\int_{0}^{1}-(x^3-x)\,dx$ (correct answer)
$\displaystyle \int_{-1}^{1}-(x^3-x)\,dx$
$\displaystyle \int_{-1}^{1}\big((x^3-x)+0\big)\,dx$
$\displaystyle \int_{-1}^{1}\left(0-(x^3-x)\right)dx$

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect multiple times. The curve y = x³ - x intersects y = 0 at x = -1, 0, 1, dividing [-1, 1] into subintervals. In [-1, 0], the cubic is above the x-axis, as at x = -0.5, (-0.5)³ - (-0.5) = 0.375 > 0. In [0, 1], it is below, as at x = 0.5, (0.5)³ - 0.5 = -0.375 < 0, necessitating a split for positive area. A tempting distractor like choice A fails because it computes net signed area, allowing positive and negative regions to cancel. To handle such problems generally, always solve for intersection points, test a point in each subinterval to determine the upper function, and integrate the difference accordingly in each piece.

Question 13

For $y=x^2-1$ and $y=0$ on $[-2,2]$ , which setup gives the total area between curves?

$\displaystyle \int_{-2}^{2}(x^2-1)\,dx$
$\displaystyle \int_{-2}^{2}(1-x^2)\,dx$
$\displaystyle \int_{-2}^{-1}(x^2-1)dx+\int_{-1}^{1}(1-x^2)dx+\int_{1}^{2}(x^2-1)dx$ (correct answer)
$\displaystyle \int_{-2}^{2}-(x^2-1)\,dx$
$\displaystyle \int_{-2}^{2}\left((x^2-1)+0\right)dx$

Explanation: This problem requires multi-interval area reasoning to compute the total area between curves that intersect. The interval must be split at x=-1 and x=1 because those are where y=x^2-1 crosses y=0 within [-2,2]. From -2 to -1, y=x^2-1 is above y=0 since x^2-1>0 for |x|>1. From -1 to 1, y=0 is above y=x^2-1 since x^2-1<0. From 1 to 2, y=x^2-1 is above y=0 again. One tempting distractor, such as choice A, fails because it computes the net area, allowing positive and negative regions to cancel out. A transferable strategy is to always locate all intersection points, divide the interval accordingly, and in each subinterval integrate the absolute difference by subtracting the bottom function from the top function.

Question 14

A ball’s height is $h(t)=16t-4t^2$ for $0\le t\le 4$ . What is the average height on $[0,4]$ ?

$\int_0^4 h(t)\,dt$
$h(2)$
$\dfrac{h(0)+h(4)}{2}$
$\dfrac{1}{4}\int_0^4 h(t)\,dt$
$\dfrac{32}{3}$ (correct answer)

Explanation: This question asks for the average value of the height function h(t). The average value of h(t) = 16t - 4t² on [0,4] is (1/4)∫[0 to 4] (16t - 4t²)dt = (1/4)[8t² - 4t³/3] from 0 to 4 = (1/4)(128 - 256/3) = (1/4)(128/3) = 32/3. Choice D shows the correct formula setup but doesn't evaluate it. The correct answer is the numerical result E = 32/3.

Question 15

Given $y^2\sin x + x\cos y=2$ , what is $\dfrac{dy}{dx}$ at a general point $(x,y)$ ?

$\dfrac{dy}{dx}=\dfrac{-y^2\cos x-\cos y}{2y\sin x-x\sin y}$ (correct answer)
$\dfrac{dy}{dx}=\dfrac{-y^2\cos x-\cos y}{2y\sin x+x\sin y}$
$\dfrac{dy}{dx}=\dfrac{-y^2\cos x+\cos y}{2y\sin x-x\sin y}$
$\dfrac{dy}{dx}=\dfrac{-y^2\cos x-\cos y}{2y\sin x}$
$\dfrac{dy}{dx}=\dfrac{-y^2\cos x}{2y\sin x-x\sin y}$

Explanation: This problem involves implicit differentiation of y²sin x + x cos y = 2. For y²sin x, we use the product rule: y²·cos x + sin x·2y·dy/dx. For x cos y, we get x·(-sin y)·dy/dx + cos y·1. Setting up: y²cos x + 2y sin x·dy/dx - x sin y·dy/dx + cos y = 0. Collecting dy/dx terms: dy/dx(2y sin x - x sin y) = -y²cos x - cos y. Therefore, dy/dx = (-y²cos x - cos y)/(2y sin x - x sin y). Choice B has a plus sign in the denominator (2y sin x + x sin y), which is incorrect—the negative sign comes from differentiating cos y. Remember: when differentiating products involving both x and y, apply the product rule carefully and track the signs from derivatives of trigonometric functions.

Question 16

A function $J$ is continuous on $[-5,-1]$ with $J(-5)=6$ and $J(-1)=2$ . Does the Mean Value Theorem guarantee a $c$ with $J'(c)=-1$ ?

Yes, because $J$ is continuous on $[-5,-1]$ and differentiable on $(-5,-1)$ , so some $c$ has $J'(c)=\dfrac{2-6}{-1-(-5)}=-1$ . (correct answer)
No, because the interval endpoints are negative.
Yes, because $J$ decreases by 4.
No, because MVT requires $J(-5)=J(-1)$ .
Yes, because $J$ is continuous, so $J'(c)$ equals $-4$ .

Explanation: The Mean Value Theorem requires J to be continuous on [-5,-1] and differentiable on (-5,-1). Since these conditions are satisfied, MVT guarantees there exists c in (-5,-1) where J'(c) equals the average rate of change. Computing: (J(-1)-J(-5))/(-1-(-5)) = (2-6)/4 = -4/4 = -1. Therefore, MVT does guarantee a point where J'(c) = -1. A common misconception is that negative x-values prevent MVT application, but the theorem works on any interval where continuity and differentiability conditions are met. The location of the interval on the x-axis is irrelevant to MVT's validity.

Question 17

During a test, $f(x)=\arccos(4x)$ gives an angle reading. What is $f'(x)$ ?

$\dfrac{4}{\sqrt{1-(4x)^2}}$
$-\dfrac{4}{\sqrt{1-(4x)^2}}$ (correct answer)
$-\dfrac{1}{\sqrt{1-(4x)^2}}$
$-\dfrac{4}{\sqrt{1+(4x)^2}}$
$-\sin(4x)$

Explanation: This requires differentiating arccos with a composite argument. The derivative of arccos(u) is -1/√(1-u²), noting the negative sign that distinguishes it from arcsin. With u = 4x and the chain rule: f'(x) = -1/√(1-(4x)²) · d/dx(4x) = -1/√(1-(4x)²) · 4 = -4/√(1-(4x)²). Option A incorrectly has a positive sign, forgetting that arccos has a negative derivative unlike arcsin. Remember the mnemonic: arccos goes down (negative derivative) while arcsin goes up (positive derivative).

Question 18

A bank account satisfies $\frac{dA}{dt}=rA$ with $r\neq 0$ . Which is the general solution?

$A(t)=rt+C$
$A(t)=Ce^{rt}$ (correct answer)
$A(t)=Ce^{-rt}$
$A(t)=C+e^{rt}$
$A(t)=C-rt$

Explanation: This problem involves solving an exponential differential equation representing compound interest. When we have $\frac{dA}{dt}=rA$ with $r\neq 0$ , the account balance changes at a rate proportional to its current amount. This proportional relationship characterizes compound interest, where larger balances earn proportionally more (if $r>0$ ) or lose proportionally more (if $r<0$ ). The standard solution to $\frac{dy}{dt} = ky$ is $y = Ce^{kt}$ , so we get $A(t) = Ce^{rt}$ . Choice A represents simple interest (linear growth), but compound interest requires the exponential relationship from proportional rates. To recognize compound interest models: when the rate of change is proportional to the current balance ( $\frac{dA}{dt} = rA$ ), the solution is always exponential.

Question 19

Which definite integral matches $\sum_{m=1}^{20} \cos\!\left(\pi+\frac{m\pi}{10}\right)\left(\frac{\pi}{10}\right)$ ?

$\displaystyle \int_{0}^{2\pi} \cos x\,dx$
$\displaystyle \int_{\pi}^{3\pi} \cos x\,dx$ (correct answer)
$\displaystyle \int_{\pi}^{3\pi} \cos\!\left(\pi+\frac{x\pi}{10}\right)\,dx$
$\displaystyle \int_{\pi}^{3\pi} \cos x\left(\frac{\pi}{10}\right)\,dx$
$\displaystyle \int_{\pi}^{2\pi} \cos x\,dx$

Explanation: This problem requires converting $\sum_{m=1}^{20} \cos\!\left(\pi+\frac{m\pi}{10}\right)\left(\frac{\pi}{10}\right)$ to integral form. The argument $\pi+\frac{m\pi}{10}$ tells us the $x$ -values where we evaluate cosine. As $m$ goes from 1 to 20, these values range from $\pi+\frac{\pi}{10} = \frac{11\pi}{10}$ to $\pi+\frac{20\pi}{10} = 3\pi$ . With $\Delta x = \frac{\pi}{10}$ and 20 rectangles, the interval spans $20 \times \frac{\pi}{10} = 2\pi$ , so we integrate from $\frac{11\pi}{10} - \frac{\pi}{10} = \pi$ to $3\pi$ . The integrand is simply $\cos x$ evaluated at these points. Choice C incorrectly keeps the linear expression inside the cosine function, not recognizing that $\pi+\frac{m\pi}{10}$ specifies where to evaluate cosine. To convert any trigonometric Riemann sum, identify the evaluation points from the argument expression and integrate the base trig function over that domain.

Question 20

If $x^2+\cos y=xy$ , what is $\dfrac{dy}{dx}$ in terms of $x$ and $y$ ?

$\dfrac{y-2x}{x+\sin y}$ (correct answer)
$\dfrac{2x-y}{x+\sin y}$
$\dfrac{y-2x}{x-\sin y}$
$\dfrac{y-2x}{\sin y}$
$\dfrac{2x}{y}$

Explanation: This problem uses implicit differentiation on x² + cos(y) = xy. Differentiating gives 2x - sin(y)·dy/dx = x·dy/dx + y, where the chain rule is applied to cos(y) and the product rule to xy. Rearranging to collect dy/dx terms yields -sin(y)·dy/dx - x·dy/dx = y - 2x, which factors as dy/dx(-sin(y) - x) = y - 2x. Since -sin(y) - x = -(x + sin(y)), we get dy/dx = (y - 2x)/(x + sin(y)). Choice B has the numerator reversed as 2x - y, which would result from a sign error when moving the 2x term. The recognition strategy is to carefully track negative signs, especially with trigonometric derivatives.

Question 21

A cyclist’s power output is $p(t)$ watts at time $t$ seconds. What does $\int_{0}^{120} p(t)\,dt$ represent?

The cyclist’s power output at $t=120$ , in watts
The total energy expended from $t=0$ to $t=120$ , in watt-seconds (joules) (correct answer)
The average power output from $t=0$ to $t=120$ , in watt-seconds
The total time ridden from $t=0$ to $t=120$ , in seconds
The net change in power output from $t=0$ to $t=120$ , in watts

Explanation: This problem tests interpreting definite integrals as accumulation in applied contexts. Since p(t) represents power output in watts (joules per second), the definite integral ∫₀¹²⁰ p(t)dt accumulates power over time to give total energy expended. The units verify this: watts × seconds = watt-seconds = joules, confirming the integral represents total energy from t=0 to t=120. Choice C incorrectly suggests average power in watt-seconds, but average power would have units of watts, not watt-seconds. When power (energy/time) is integrated over time, the result is total energy—a key principle in physics applications.

Question 22

For $U(x)=\begin{cases}x^2+2,&x\ne-3\\11,&x=-3\end{cases}$ , is $U$ continuous at $x=-3$ , and why?

Yes; $\lim_{x\to-3}U(x)=11$ and $U(-3)=11$ . (correct answer)
No; $\lim_{x\to-3}U(x)=9$ but $U(-3)=11$ .
No; $\lim_{x\to-3}U(x)$ does not exist because it is quadratic.
No; $U(-3)$ is undefined.
Yes; the limit exists, so continuity holds even if $U(-3)$ differed.

Explanation: Continuity at a point requires that the function is defined at that point, the limit exists as x approaches that point, and the limit equals the function's value there. For U(x) at x = -3, U(-3) = 11 is defined, and the limit is lim (x² + 2) = 11, which matches U(-3), so it is continuous. A common omission is forgetting to check if the limit equals the redefined value at the point, as the piecewise definition might suggest a hole otherwise. Here, since the limit of the quadratic equals the given value, continuity holds. Remember, redefining a function at a single point can make it continuous if it fills the hole correctly. To verify continuity at any point a, use this checklist: ensure f(a) is defined, compute lim_{x→a} f(x) and confirm it exists, then check if lim_{x→a} f(x) = f(a).

Question 23

If $w(x)=\cot(\pi x)$ , what is $w'(x)$ ?

$\pi\csc^2(\pi x)$
$-\csc^2(\pi x)$
$-\pi\csc^2(\pi x)$ (correct answer)
$-\pi\sec^2(\pi x)$
$\pi\sec^2(\pi x)$

Explanation: This problem requires differentiating cotangent with a linear argument involving π, testing reciprocal trig derivatives and chain rule precision. The derivative of cot(u) is -csc²(u)·u', and for w(x) = cot(πx), we have u = πx, giving u' = π. Thus w'(x) = -csc²(πx)·π = -π·csc²(πx). A common error would be confusing cotangent's derivative with tangent's and writing π·sec²(πx), but cotangent derivatives always involve cosecant squared, not secant squared. Remember to treat π as a constant coefficient when applying the chain rule, just like any other numerical coefficient.

Question 24

A differentiable function $L(x)$ gives light intensity. Which expression represents the slope of $L$ at $x=1$ ?

$L(1)$
$\dfrac{L(2)-L(1)}{1}$
$L'(1)$ (correct answer)
$\dfrac{L(1)-L(0)}{1}$
$\dfrac{d}{dx}(1)$

Explanation: This question requires interpreting derivative notation in the context of light intensity slope. When $L(x)$ gives light intensity as a function of some variable $x$ , the slope of $L$ at $x=1$ represents the derivative $L'(1)$ . This notation shows the instantaneous rate of change of light intensity with respect to $x$ at that point, which geometrically corresponds to the slope of the tangent line to the graph of $L$ at $x=1$ . The slope of a function at a point is precisely its derivative at that point. Choice B represents the average rate of change between $x=1$ and $x=2$ , which gives the slope of a secant line, not the slope of the function itself at $x=1$ . For slopes of curves (functions), use derivative notation to get the instantaneous slope rather than average rates.

Question 25

A differentiable $f$ has $f'(1)=0$ and $f''(1)>0$ ; classify the point $x=1$ .

Local minimum at $x=1$ . (correct answer)
Local maximum at $x=1$ .
Inflection point at $x=1$ .
Cannot be determined from the information given.
Neither a maximum nor minimum because $f''(1)>0$ .

Explanation: The Second Derivative Test uses concavity information to classify the type of critical point present. With $f'(1)=0$ confirming a critical point and $f''(1)>0$ indicating positive concavity, the function exhibits upward curvature at $x=1$ . Upward concavity at a critical point creates a valley-like shape, establishing a local minimum. Choice B might mislead students who link positive direction with maximums, but positive second derivative specifically describes the upward curvature of minimums. Remember to confirm both critical point existence and definitive second derivative sign before classification.

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