AP Calculus AB

AP Calculus AB Practice Test: Practice Test 5

Practice Test 5 for AP Calculus AB: real questions and explanations from the Varsity Tutors practice-test pool.

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Question 1 of 25

A rectangle has area $A=x(18-x)$ for $0<x<18$ ; which $x$ maximizes $A$ ?

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Question 1

A rectangle has area $A=x(18-x)$ for $0<x<18$ ; which $x$ maximizes $A$ ?

$x=18$
$x=0$
$x=6$
$x=9$ (correct answer)
$x=12$

Explanation: This problem requires solving optimization problems using calculus, a key skill in AP Calculus AB. To maximize the area A = x(18 - x), compute the derivative A' = 18 - 2x and set it to zero, yielding a critical point at x = 9. Evaluating at x = 9 gives the maximum, while endpoints yield zero. The second derivative A'' = -2 is negative, confirming a maximum. A tempting distractor is x = 6, but it yields a smaller area as it is not the critical point. To solve optimization problems, express the quantity as a function of one variable, find critical points by setting the derivative to zero, and evaluate at critical points and endpoints to find the maximum or minimum.

Question 2

Given $g'(5)=0$ and $g''(5)>0$ , what does the Second Derivative Test conclude at $x=5$ ?

Local minimum at $x=5$ . (correct answer)
Local maximum at $x=5$ .
Inflection point at $x=5$ .
Cannot be determined because $g'(5)=0$ .
No extremum because $g''(5)>0$ .

Explanation: The Second Derivative Test determines the nature of critical points by analyzing the concavity at those locations. With $g'(5)=0$ confirming a critical point and $g''(5)>0$ indicating positive concavity, the graph is concave up at $x=5$ . Concave up curvature at a critical point forms a valley, establishing a local minimum. Choice B could mislead students who associate positive values with maximums, but positive second derivative specifically indicates upward concavity (minimum). The test is applicable when both the critical point condition and second derivative sign are clearly determined.

Question 3

A differentiable function $F$ has $F(6)=1$ and $F'(6)=3$ . Approximate $F(5.9)$ using local linearity.

$0.7$ (correct answer)
$1.3$
$0.97$
$3.9$
$1.03$

Explanation: This problem uses linearization for local estimation. Given F(6) = 1 and F'(6) = 3, F(5.9) ≈ 1 + 3(-0.1) = 0.7. Conceptually, it's the tangent line extension. Wrong setup: Positive Δx when decreasing. Some subtract without sign check. Strategy: Compute Δx = x - a, multiply, add to f(a).

Question 4

If $f(x)=\frac{x^2+1}{3}$ , find $\lim_{x\to -2} f(x)$ using limit laws.

$\frac{5}{3}$ (correct answer)
$\frac{1}{3}$
$\frac{4}{3}$
$\frac{7}{3}$
$\frac{5}{2}$

Explanation: This rational function with a quadratic numerator requires careful application of limit laws. Since the denominator is non-zero at $x = -2$ , we can use direct substitution: $\lim_{x\to -2} f(x) = \frac{(-2)^2 + 1}{3} = \frac{4 + 1}{3} = \frac{5}{3}$ . We apply the quotient rule, with the numerator limit being $(-2)^2 + 1 = 5$ and denominator limit being 3. A common error would be sign mistakes when substituting negative values or incorrectly computing $(-2)^2$ . The key strategy is to carefully substitute negative values and apply order of operations correctly in rational functions.

Question 5

Compute $\int (3x^2-4x)(x^3-2x^2+7)^5\,dx$ from a composite response function.

$\dfrac{(x^3-2x^2+7)^6}{6}+C$ (correct answer)
$\dfrac{(x^3-2x^2+7)^6}{3}+C$
$(x^3-2x^2+7)^6+C$
$\dfrac{(3x^2-4x)(x^3-2x^2+7)^6}{6}+C$
$\dfrac{(x^3-2x+7)^6}{6}+C$

Explanation: The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is x³ - 2x² + 7, whose derivative 3x² - 4x exactly matches the factor outside. Set u = x³ - 2x² + 7, so du = (3x² - 4x) dx, transforming the integral into ∫ u⁵ du. Integrating gives (1/6) u⁶ + C, or (x³ - 2x² + 7)⁶ / 6 + C. A tempting distractor like (x³ - 2x² + 7)⁶ / 3 + C fails because it uses the wrong divisor, possibly confusing the exponent with the coefficient. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.

Question 6

Direct substitution into the limit $\lim_{x \to 0} \frac{e^x - 1}{\sin(x)}$ yields $\frac{0}{0}$ . Which procedure is most effective for evaluating this limit without using L'Hôpital's Rule?

Multiply the numerator and denominator by $x$ and rearrange to use the known limits for $\frac{e^x-1}{x}$ and $\frac{x}{\sin(x)}$ . (correct answer)
Multiply the numerator and denominator by the conjugate of the numerator, $e^x + 1$ , to simplify the expression.
Apply the Squeeze Theorem by bounding the function between two simpler functions that approach the same value.
Factor $e^x$ from the numerator and use trigonometric identities to simplify the denominator.

Explanation: This limit can be resolved by using two known fundamental limits: $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ and $\lim_{x \to 0} \frac{\sin x}{x} = 1$ . The expression can be rewritten as $\lim_{x \to 0} \frac{(e^x - 1)/x}{\sin(x)/x}$ . Using the quotient property of limits, the limit is $\frac{\lim_{x \to 0} \frac{e^x-1}{x}}{\lim_{x \to 0} \frac{\sin x}{x}} = \frac{1}{1} = 1$ .

Question 7

If $T(x)=\int_{-3}^{x}\left(e^{2t}-\cos t\right)dt$ , what is $T'(x)$ ?

$2e^{2x}+\sin x$
$\displaystyle \int_{-3}^{x}\left(e^{2t}-\cos t\right)dt$
$e^{2x}-\cos x$ (correct answer)
$\displaystyle \int_{-3}^{x}\left(2e^{2t}+\sin t\right)dt$
$\displaystyle (e^{2t}-\cos t)\big|_{t=-3}$

Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1, where the derivative of an integral with variable upper limit equals the integrand evaluated at that limit. Given $T(x) = \int_{-3}^{x}(e^{2t} - \cos t)\,dt$ , we get $T'(x) = e^{2x} - \cos x$ by replacing t with x in the integrand. The process is purely evaluation; no integration or differentiation of the integrand is performed. Choice A ( $2e^{2x} + \sin x$ ) shows the derivatives of $e^{2x}$ and $-\cos x$ , which would result from incorrectly differentiating after applying FTC Part 1. To master this concept, remember that FTC Part 1 converts "derivative of integral" problems into simple substitution exercises.

Question 8

Let $X(x)=\begin{cases}2x+3,&x< -1\\1,&x=-1\\2x+3,&x>-1\end{cases}$ . Is $X$ continuous at $x=-1$ , and why?

Yes; $\lim_{x\to-1}X(x)=1$ exists and equals $X(-1)=1$ . (correct answer)
No; $\lim_{x\to-1}X(x)=1$ exists but $X(-1)=5$ .
No; $\lim_{x\to-1}X(x)=1$ exists but $X(-1)=1$ is undefined.
No; $\lim_{x\to-1}X(x)$ does not exist because one-sided limits differ.
No; $\lim_{x\to-1}X(x)=0$ but $X(-1)=1$ .

Explanation: X at -1: X(-1)=1, limit 2x+3=1 matches. Continuous. Omission: thinking middle differs. Checklist: f(a), limit, equality.

Question 9

Let $\theta(x)=\begin{cases}x+5,&x< -1\\x^2,&x\ge -1\end{cases}$ . What type of discontinuity does $\theta$ have at $x=-1$ ?

Removable discontinuity
Infinite discontinuity
Jump discontinuity (correct answer)
No discontinuity (continuous)
Oscillating discontinuity

Explanation: This function has a jump discontinuity at x = -1. The left-hand limit is lim(x→-1⁻) (x+5) = 4, while the right-hand limit is lim(x→-1⁺) x² = 1, and θ(-1) = (-1)² = 1. Since both one-sided limits exist but are unequal (4 ≠ 1), this creates a jump discontinuity. The function jumps from approaching 4 on the left to the value 1 at and to the right of x = -1. Students might think this is continuous because θ(-1) equals the right-hand limit, but continuity requires both one-sided limits to be equal. To identify jump discontinuities: verify that both one-sided limits exist but have different values.

Question 10

A graph of $f'$ touches the $x$ -axis at $x=3$ and stays positive on both sides; which could be $f$ at $x=3$ ?

A local maximum at $x=3$ .
A local minimum at $x=3$ .
A horizontal tangent at $x=3$ but no local extremum there. (correct answer)
A discontinuity at $x=3$ .
A vertical tangent at $x=3$ .

Explanation: This problem explores a special case where the derivative touches but doesn't cross the x-axis. When f' touches the x-axis at x = 3 and stays positive on both sides, it means f'(3) = 0 but f' doesn't change sign. Since f' remains positive near x = 3, the function f continues to increase through this point. The horizontal tangent at x = 3 (due to f'(3) = 0) combined with f increasing on both sides means there's neither a local maximum nor minimum—just a horizontal tangent. Choice B incorrectly identifies this as a local minimum, which would require f' to change from negative to positive. To identify extrema, always check whether f' actually changes sign at critical points; a mere touch of the x-axis without sign change indicates a horizontal tangent with no extremum.

Question 11

A lake’s pollutant concentration $C(t)$ decreases at a rate proportional to $C(t)$ due to natural cleanup. Which differential equation models $C$ ?

$\dfrac{dC}{dt}=\dfrac{k}{C}$
$\dfrac{dC}{dt}=-kC$ (correct answer)
$\dfrac{dC}{dt}=kC$
$\dfrac{dt}{dC}=-kC$
$\dfrac{dC}{dt}=-k(C-1)$

Explanation: This problem models natural cleanup processes where pollutant concentration decreases at a rate proportional to the current concentration. The lake's "natural cleanup" causes the concentration to decrease proportionally to C(t), giving dC/dt = -kC where k > 0. The negative sign indicates decreasing concentration as the lake's natural processes remove pollutants. Choice C would represent increasing pollution rather than cleanup. Choice E introduces an unnecessary constant term that would represent a threshold concentration. Environmental cleanup problems typically follow exponential decay where the cleanup rate depends on how much pollution remains.

Question 12

A continuous function $p$ on $[1,5]$ has critical points $x=2,4$ . If $p(1)=0,p(2)=3,p(4)=1,p(5)=2$ , where is the absolute maximum?

At $x=5$
At $x=2$ (correct answer)
At $x=4$
At $x=1$
At $x=1$ or $x=5$ only

Explanation: This problem assesses the Candidates Test, which is used to find absolute extrema of a continuous function on a closed interval. To apply the Candidates Test, we evaluate the function at all critical points inside the interval and at the endpoints. Critical points are where the derivative is zero or undefined, potentially indicating local maxima or minima. Endpoints must be checked because the absolute extremum could occur there, even without a zero derivative. A tempting distractor might be choosing x=4, but p(4)=1 is less than p(2)=3, so it's not the maximum. Remember, the checklist for candidates is: list all critical points in the interval, include both endpoints, evaluate f at each, and compare the values to find the max or min.

Question 13

Water flows into a tank at rate $r(t)=3t-6$ liters/min for $0\le t\le4$ . What is the net change in volume?

$0$ liters (correct answer)
$12$ liters
$-12$ liters
$6$ liters
$-6$ liters

Explanation: This problem requires using accumulation reasoning to find the net change in water volume from the given rate function. The net change equals the integral of r(t) = 3t - 6 from 0 to 4, which represents the total accumulation over the time interval. Computing ∫(3t - 6)dt = (3t²/2 - 6t) evaluated from 0 to 4 gives (3(16)/2 - 24) - 0 = 24 - 24 = 0 liters. The rate starts negative (water flowing out), becomes zero at t = 2, then positive (water flowing in), with equal amounts flowing out and in. Choice B (12 liters) might tempt students who only consider the positive accumulation after t = 2, ignoring the negative accumulation before. When a rate function changes sign, always integrate over the entire interval to find the net accumulation, as positive and negative changes may cancel.

Question 14

A function $v$ is continuous on $[0,4]$ with $v(0)=5$ and $v(4)=1$ . Does the Mean Value Theorem guarantee a $c$ with $v'(c)=-1$ ?

No, because $v(0)\ne v(4)$ .
Yes, because $v$ is continuous on $[0,4]$ and differentiable on $(0,4)$ , so some $c$ has $v'(c)=\dfrac{1-5}{4-0}=-1$ . (correct answer)
Yes, because $v$ must cross $y=3$ .
Yes, because $v$ is decreasing on the interval.
No, because MVT only applies to polynomials.

Explanation: The Mean Value Theorem applies when v is continuous on [0,4] and differentiable on (0,4). Since both conditions are satisfied, MVT guarantees there exists c in (0,4) where v'(c) equals the average rate of change. The average rate of change is (v(4)-v(0))/(4-0) = (1-5)/4 = -4/4 = -1. Therefore, MVT does guarantee a point where v'(c) = -1. A common misconception is that MVT requires equal endpoint values - that's actually Rolle's Theorem, a special case of MVT. The general MVT works whenever the continuity and differentiability conditions are met, regardless of whether endpoints are equal.

Question 15

If $p$ has a vertical tangent at $x=5$ , does $p'(5)$ exist, and why?

Yes; vertical tangents correspond to derivative $0$ .
No; a vertical tangent indicates an infinite slope, so the derivative does not exist. (correct answer)
Yes; as long as $p$ is continuous at $x=5$ , the derivative exists.
No; a vertical tangent means the function is discontinuous at $x=5$ .
Yes; the derivative exists because one-sided limits of $p$ match.

Explanation: This question examines the link between differentiability and continuity, specifically regarding vertical tangents. A vertical tangent at x=5 indicates that the slope approaches infinity, meaning the limit of the difference quotient does not exist as a finite number. For p'(5) to exist, this limit must be finite and equal from both sides, but an infinite slope prevents that. The function can be continuous at x=5 with a vertical tangent, yet differentiability requires a finite derivative. Choice C is a tempting distractor, stating continuity ensures the derivative exists, but it fails because vertical tangents make the derivative undefined despite continuity. To check differentiability at a point, verify continuity there and ensure the left- and right-hand derivatives exist and are equal.

Question 16

A slope field corresponds to $\frac{dy}{dx}=\frac{y-x}{y+x}$ . At points where $y=x$ , slope segments are

vertical.
horizontal. (correct answer)
always negative.
undefined.
equal to $1$ .

Explanation: This question uses slope field reasoning about slopes along specific curves. For $\frac{dy}{dx} = \frac{y-x}{y+x}$ , at points where $y = x$ , we have $\frac{dy}{dx} = \frac{x-x}{x+x} = \frac{0}{2x} = 0$ (assuming $x ≠ 0$ ). When $y = x$ and $x ≠ 0$ , the numerator becomes zero while the denominator remains non-zero, making the slope zero. This means slope segments are horizontal along the line $y = x$ (except at the origin where both numerator and denominator are zero). Choice A fails because slopes are finite (zero), not infinite, along $y = x$ . To find slope behavior along specific curves in slope fields, substitute the curve's equation into the differential equation and simplify.

Question 17

A population of bacteria is modeled by the logistic function $P(t) = \frac{2000}{1 + 49e^{-0.2t}}$ , where $t$ is time in days. What number does the population approach as time increases without bound?

$49$
$2000$ (correct answer)
$50$
The population grows infinitely.

Explanation: To find the long-term behavior of the population, we need to evaluate the limit of $P(t)$ as $t \to \infty$ . As $t \to \infty$ , the term $-0.2t$ approaches $-\infty$ , so $e^{-0.2t}$ approaches 0. The limit is then $\lim_{t \to \infty} \frac{2000}{1 + 49e^{-0.2t}} = \frac{2000}{1 + 49(0)} = 2000$ . This value is the carrying capacity of the population.

Question 18

From the graph, $H(x)$ approaches $-2$ as $x$ approaches $-1$ from the right only. Which limit notation represents this?

$\displaystyle \lim_{x\to -1^+} H(x)=-2$ (correct answer)
$\displaystyle \lim_{x\to -1} H(x)=-2$
$\displaystyle H(-1)=-2$
$\displaystyle \lim_{x\to -1^-} H(x)=-2$
$\displaystyle \lim_{x\to -2} H(x)=-1$

Explanation: Graph H(x) to -2 from right at -1. \lim_{x\to -1^+} H(x)=-2 correct. Valid right. Error: \lim_{x\to -1} H(x)=-2. H(-1)=-2 value. Specify. Transferable notation checklist: 1. Use 'lim' for limits. 2. Specify the approach with x\to a. 3. Add ^+ or ^- for one-sided limits if needed. 4. Ensure the expression equals the approached value.

Question 19

A probability density uses $\int \dfrac{10x}{\sqrt{5x^2+9}}\,dx$ . Find an antiderivative.

$2\sqrt{5x^2+9}+C$ (correct answer)
$\sqrt{5x^2+9}+C$
$\dfrac{10x}{\sqrt{5x^2+9}}+C$
$5\sqrt{5x^2+9}+C$
$2\sqrt{5x+9}+C$

Explanation: The skill here is integration using u-substitution, which simplifies the integral by replacing a composite function with a single variable. Recognize that the inner function is 5x² + 9, whose derivative 10x exactly matches the numerator. Set u = 5x² + 9, so du = 10x dx, transforming the integral into ∫ du / √u. Integrating gives 2 √u + C, or 2 √(5x² + 9) + C. A tempting distractor like √(5x² + 9) + C fails because it misses the factor of 2 from integrating u^{-1/2}, which should be 2 u^{1/2}. To recognize opportunities for u-substitution, look for an inner function whose derivative is a constant multiple of the remaining factors in the integrand.

Question 20

For $r(x)=\begin{cases}3,&x<0\\1,&x=0\\-3,&x>0\end{cases}$ , what type of discontinuity occurs at $x=0$ ?

Infinite discontinuity
No discontinuity (continuous at $x=0$ )
Removable discontinuity
Jump discontinuity (correct answer)
Oscillating discontinuity

Explanation: For the piecewise function r(x), we examine the behavior at x=0. The left limit is lim(x→0⁻) r(x) = 3, and the right limit is lim(x→0⁺) r(x) = -3. Since these one-sided limits are finite but unequal (3 ≠ -3), the function has a jump discontinuity at x=0. The fact that r(0) = 1 is defined doesn't affect the discontinuity type—what matters is that the left and right limits differ. Students sometimes focus too much on the function value at the point, but for jump discontinuities, the key is the mismatch between one-sided limits. Classification strategy: when one-sided limits are both finite but different, it's always a jump discontinuity regardless of the function's value at that point.

Question 21

A runner’s distance $s(t)$ increases with speed proportional to $\sqrt{s(t)}$ . Which differential equation models $s$ ?

$\dfrac{ds}{dt}=k\sqrt{s}$ (correct answer)
$\dfrac{ds}{dt}=ks$
$\dfrac{ds}{dt}=\dfrac{k}{\sqrt{s}}$
$\dfrac{ds}{dt}=k\sqrt{t}$
$\dfrac{dt}{ds}=k\sqrt{s}$

Explanation: This problem describes motion where speed increases with a square root dependency on current distance traveled. The distance "increases with speed proportional to $\sqrt{s(t)}$ ," giving $ds/dt = k\sqrt{s}$ where $k > 0$ . The positive sign indicates increasing distance, and the square root relationship might represent physical effects like momentum or energy relationships. Choice B would be exponential growth in distance. Choice C has an inverse relationship that would slow down with more distance. Square root speed relationships can arise in problems involving energy conservation or acceleration effects.

Question 22

Select the washer-method integral when the region between $y=\tfrac{3}{2}x+2$ and $y=\tfrac{1}{2}x+1$ on $[0,4]$ is revolved about the x-axis.

$\pi\int_{0}^{4}\left[\left(\tfrac{1}{2}x+1\right)^2-\left(\tfrac{3}{2}x+2\right)^2\right]dx$
$\pi\int_{0}^{4}\left[\left(\tfrac{3}{2}x+2\right)^2-\left(\tfrac{1}{2}x+1\right)^2\right]dx$ (correct answer)
$\pi\int_{0}^{4}\left(\tfrac{3}{2}x+2\right)^2dx$
$\pi\int_{0}^{4}\left[\left(\tfrac{3}{2}x+2\right)-\left(\tfrac{1}{2}x+1\right)\right]^2dx$
$\pi\int_{0}^{4}\left[\left(\tfrac{3}{2}x+2\right)^2+\left(\tfrac{1}{2}x+1\right)^2\right]dx$

Explanation: The washer method calculates volumes when revolving regions between curves around an axis, creating washers with outer and inner radii. Here, revolving around the x-axis, the radii are the y-values of the curves. Compare the functions: (3/2)x + 2 > (1/2)x + 1 on [0,4], so outer radius is (3/2)x + 2, inner is (1/2)x + 1. Thus, the integral is π ∫ from 0 to 4 of [((3/2)x + 2)^2 - ((1/2)x + 1)^2] dx. A tempting distractor is choice D, which squares the difference, erroneously simplifying the washer to a single radius. For any washer method setup, checklist: confirm the axis, identify outer and inner functions relative to the axis, ensure outer > inner, set up π ∫ (outer^2 - inner^2) d(variable perpendicular to axis), with limits where the region exists.

Question 23

Let $E(x)=\int_{-2}^{x}\dfrac{t^2+1}{t^2+4}\,dt$ . What is $E'(x)$ ?

$\dfrac{x^2+1}{x^2+4}$ (correct answer)
$\dfrac{2x(x^2+4)-2x(x^2+1)}{(x^2+4)^2}$
$\int_{-2}^{x}\dfrac{t^2+1}{t^2+4}\,dt$
$\dfrac{t^2+1}{t^2+4}$
$\left[\int \dfrac{t^2+1}{t^2+4}\,dt\right]_{-2}^{x}$

Explanation: This problem requires applying the Fundamental Theorem of Calculus Part 1 to find the derivative of an accumulation function. The function E(x) is defined as the integral from -2 to x of (t² + 1)/(t² + 4) dt, which accumulates the area under the curve of f(t) = (t² + 1)/(t² + 4) from a fixed lower limit to the variable upper limit x. According to FTC1, the derivative E'(x) equals the integrand evaluated at x, which is (x² + 1)/(x² + 4). This result holds because the instantaneous rate of change of the accumulated integral at x is precisely the value of the function being integrated at that point. A tempting distractor is choice E, which represents the integral of (t² + 1)/(t² + 4) dt evaluated from -2 to x, but that actually computes E(x) itself, not its derivative. To recognize FTC Part 1 problems, look for functions defined as definite integrals with a variable upper limit and a fixed lower limit, and questions asking for the derivative of that function.

Question 24

A function is $B(x)=\sin x+\cos x+\tan x$ . Which differentiation procedure is sufficient to find $B'(x)$ ?

Sum rule with basic trig derivatives (correct answer)
Product rule and chain rule
Quotient rule
Implicit differentiation
Logarithmic differentiation

Explanation: This problem involves selecting the appropriate differentiation procedure for $B(x)=\sin x+\cos x+\tan x$ . This function is a sum of three trigonometric functions, so the sum rule applies - we differentiate each term separately using basic trigonometric derivatives. The derivative is $\cos x - \sin x + \sec^2 x$ . No complex rules like product rule, chain rule, quotient rule, or implicit differentiation are needed since each term is a basic trigonometric function of $x$ . This is a straightforward application of the sum rule combined with knowledge of basic trigonometric derivatives. When you encounter sums of basic functions, the sum rule with appropriate basic derivatives is sufficient.

Question 25

Let $k(x)=\begin{cases}x^2,&x<2\\4,&x=2\\5,&x>2\end{cases}$ . Is $k$ continuous at $x=2$ , and why?

No; $\lim_{x\to2^-}k(x)=4$ and $\lim_{x\to2^+}k(x)=5$ , so the limit does not exist. (correct answer)
Yes; $\lim_{x\to2}k(x)=4$ and $k(2)=4$ .
No; $\lim_{x\to2}k(x)=5$ but $k(2)=4$ .
No; $k(2)$ is undefined.
Yes; the right-hand limit equals $5$ , so it is continuous.

Explanation: For continuity, the limit must exist and equal f(a), with f(a) defined. At x = 2 for k(x), k(2) = 4, but left limit 4 and right 5 differ, so limit DNE, discontinuous. Often, students forget to check one-sided agreement in piecewise constants. This is a jump discontinuity. Matching pieces are crucial. To check 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).

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