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AP Calculus AB : Functions, Graphs, and Limits

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Example Questions

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Example Question #3 : Estimating Limits From Graphs Or Tables

f(x) Discgraph

True or false: In the above graph of f(x) , the value of $\lim_{x\rightarrow -4}f(x)$ is 3.

Possible Answers:

True: The removable discontinuity does not affect the limit, and the right and left limits evaluate to 3.

False: We can't take the limit where the function isn't defined.

False: The left and right limits exist, but neither of them is three.

False: The left and right limits exist, but exactly one of them isn't three.

False: Both the left and right limits are three, but the regular limit isn't.

Correct answer:

True: The removable discontinuity does not affect the limit, and the right and left limits evaluate to 3.

Explanation:

Removable discontinuities don't affect the limiting process. The limit process is essentially saying "As you get arbitrarily close to x = -4 , the function is getting arbitrarily close to ." As you can see, whether or not the function is defined at is irrelevant, because we're want to look at values close to it, but never exactly at .

Graphically, we can see that the left and right limits are both three. If you trace the graph from either the left or right of , you will end up at .

By definition, if both the right and left limit evaluate to the same thing, the actual limit must agree. It's not possible for both of them to be three, but for the limit to disagree with them (at least in a two dimensional graph!)

Thus the correct answer is that the limit exists, and is three.

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Example Question #4 : Estimating Limits From Graphs Or Tables

f(x)

Discgraph

In the above graph of f(x) , evaluate the following limits:

$\lim_{x\rightarrow 4^+} f(x)$

$\lim_{x\rightarrow 4^-} f(x)$

$\lim_{x\rightarrow 4} f(x)$

Possible Answers:

$\lim_{x\rightarrow 4^+}f(x) =5$

$\lim_{x\rightarrow 4^-}f(x) =3$

$\lim_{x\rightarrow 4}f(x) = 3$

$\lim_{x\rightarrow 4^+}f(x) =3$

$\lim_{x\rightarrow 4^-}f(x) =3$

$\lim_{x\rightarrow 4}f(x) = 3$

$\lim_{x\rightarrow 4^+}f(x) =5$

$\lim_{x\rightarrow 4^-}f(x) =3$

$\lim_{x\rightarrow 4}f(x) \: \mathrm{Does \: Not \: Exist}$

$\lim_{x\rightarrow 4^+}f(x) \: \mathrm{Does \: Not \: Exist}$

$\lim_{x\rightarrow 4^-}f(x) =3$

$\lim_{x\rightarrow 4}f(x) \: \mathrm{Does \: Not \: Exist}$

$\lim_{x\rightarrow 4^+}f(x) =3$

$\lim_{x\rightarrow 4^-}f(x) =5$

$\lim_{x\rightarrow 4}f(x) \: \mathrm{Does \: Not \: Exist}$

Correct answer:

$\lim_{x\rightarrow 4^+}f(x) =5$

$\lim_{x\rightarrow 4^-}f(x) =3$

$\lim_{x\rightarrow 4}f(x) \: \mathrm{Does \: Not \: Exist}$

Explanation:

A right limit is found graphically by starting at a point to the right of the specified x value and tracing along the graph to the left until you hit the specified x. Regardless of whether or not the function is defined, or perhaps if its defined, but not where your finger ends up, your finger should be at the limit. This is because limits don't concern themselves with what happens at the specified x value. For right limits, we're essentially asking "If I get arbitrarily close to x=4 from the right, what is f(x) getting close to?"

The same holds true for left limits, but you approach them from the other side.

In this problem, $\lim_{x\rightarrow 4^+}f(x)$ is a right limit. Following the above method, we find that as we approach x =4 from the right, we end up at . Similarly, $\lim_{x\rightarrow 4^-}f(x)$ is a left limit. Approaching x =4 from the left, we see that we arrive at .

Lastly, a regular limit is defined when both the right and left limits are defined and equal. In this case, the right limit is 5 and the left limit is 3, so the regular limit does not exist

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Example Question #1 : Understanding The Limiting Process.

Find the derivative.

y = sec (5x³)

Possible Answers:

y' = –csc(5x³)cot(5x³)(15x²)

y' = sec(5x³)tan(5x³)(15x²)

y' = –sec(5x³)tan(5x³)(15x²)

y' = –csc(5x³)cot(5x³)

y' = sec(5x³)tan(5x³)

Correct answer:

y' = sec(5x³)tan(5x³)(15x²)

Explanation:

The derivative of the function y = sec(x) is sec(x)tan(x). First take the derivative of the outside of the function: y = sec(4x³) : y' = sec(5x³)tan(5x³). Then take the derivative of the inside of the function: 5x³ becomes 15x². So your final answer is: y' = ec(5x³)tan(5x³)15x²

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Example Question #171 : Functions, Graphs, And Limits

Find the slope of the tangent line to the graph of f at x = 9, given that f(x) = –x²+ 5√(x)

Possible Answers:

18 + (5/6)

–18 – (5/6)

–18

–18 + (5/6)

Correct answer:

–18 + (5/6)

Explanation:

First find the derivative of the function.

f(x) = –x² + 5√(x)

f'(x) = –2x + 5(1/2)x^–1/2

Simplify the problem

f'(x) = –2x + (5/2x^1/2)

Plug in 9.

f'(3) = –2(9) + (5/2(9)^1/2)

= –18 + 5/(6)

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Example Question #2 : Understanding The Limiting Process.

Find the derivative

(x + 1)/(x – 1)

Possible Answers:

(–2)/(x – 1)²

(–2)/(x – 1)

(x + 1) + (x – 1)

(–2)/(x + 1)²

Correct answer:

(–2)/(x – 1)²

Explanation:

Rewrite problem.

(x + 1)/(x – 1)

Use quotient rule to solve this derivative.

((x – 1)(1) – (x + 1)(1))/(x – 1)²

(x – 1) – x – 1)/(x – 1)²

–2/(x – 1)²

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Example Question #1 : Understanding The Limiting Process.

$\frac{d}{dx}sin^{-1} (3x)$

Possible Answers:

$\frac{3}{\sqrt{1-9 x^2}}$

$\frac{3x}{\sqrt{1-9 x^2}}$

$\frac{3}{\sqrt{1-3x^2}}$

$\frac{3x}{\sqrt{1- x^2}}$

Correct answer:

$\frac{3}{\sqrt{1-9 x^2}}$

Explanation:

Use the chain rule and the formula

$\frac{d}{dx}sin^{-1}u = \frac{1}{\sqrt{1-x^2}}$

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Example Question #1 : Understanding The Limiting Process.

Find the derivative of

$f(x)=4(x^{3}-2x+11)(x-2)$

Possible Answers:

24x-2

$4x^{4}-8x^{3}-8x^{2}+60x-22$

$3x^{2}-8x+15$

$4(3x^{2}-2)$

$f'(x)=16x^{3}-24x^{2}-16x+60$

Correct answer:

$f'(x)=16x^{3}-24x^{2}-16x+60$

Explanation:

The answer is $f'(x)=16x^{3}-24x^{2}-16x+60$ . It is easy to solve if we multiply everything together first before taking the derivative.

$f(x)=4(x^{3}-2x+11)(x-2)$

$f(x)=4(x^{4}-2x^{3}+-2x^{2}+4x+11x-22)$

$f(x)=4x^{4}-8x^{3}-8x^{2}+60x-22$

$f'(x)=16x^{3}-24x^{2}-16x+60$

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Example Question #171 : Functions, Graphs, And Limits

Find $\frac{d^{2}y}{dx^{2}}$ if y^2-x^2 = 4 .

Possible Answers:

$\frac{x^2+4}{(x^2-4)^\frac{3}{2}}$

$\frac{x^2-4}{\sqrt{x^2+4}}$

$\frac{x^2}{(x^2-4)^\frac{3}{2}}$

$\frac{4}{(x^{2}+4)^{\frac{3}{2}}}$

$\frac{x}{\sqrt{4+x^2}}$

Correct answer:

$\frac{4}{(x^{2}+4)^{\frac{3}{2}}}$

Explanation:

We will have to find the first derivative of with respect to using implicit differentiation. Then, we can find $\frac{d^{2}y}{dx^{2}}$ , which is the second derivative of with respect to .

$\frac{d}{dx}[y^2-x^2]=\frac{d}{dx}[4]$

We will apply the chain rule on the left side.

$\frac{d}{dx}[y^2]\cdot \frac{dy}{dx}-\frac{d}{dx}[x^2]=0$

$2y\cdot \frac{dy}{dx}-2x=0$

We now solve for the first derivative with respect to .

$\frac{dy}{dx}=\frac{x}{y}$

In order to get the second derivative, we will differentiate $\frac{x}{y}$ with respect to . This will require us to employ the Quotient Rule.

$\frac{d}{dx}\frac{dy}{dx}=\frac{d^{2}y}{dx^{2}}=\frac{d}{dx}[\frac{x}{y}]=\frac{y\cdot \frac{d}{dx}[x]-x\cdot \frac{d}{dx}[y]}{y^{2}} =\frac{y\cdot 1-x\cdot \frac{dy}{dx}}{y^{2}}$

We will replace $\frac{\mathrm{d}y }{\mathrm{d} x}$ with $\frac{x}{y}$ .

$\frac{y-x(\frac{x}{y})}{y^{2}}=\frac{y^{2}-x^{2}}{y^{3}}$

But, from the original equation, $y^{2}-x^{2}=4$ . Also, if we solve for , we obtain $y = (4+x^{2})^{\frac{1}{2}}$ .

$\frac{d^{2}y}{dx^{2}}=\frac{y^{2}-x^{2}}{y^{3}}=\frac{4}{(x^{2}+4)^{\frac{3}{2}}}$

The answer is $\frac{4}{(x^{2}+4)^{\frac{3}{2}}}$ .

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Example Question #2 : Understanding The Limiting Process.

Differentiate f(x)=5x^3 .

Possible Answers:

15x^3

15x^2

(5/3)x^2

15x^4

8x^2

Correct answer:

15x^2

Explanation:

Using the power rule, multiply the coefficient by the power and subtract the power by 1.

f(x)=5x^3

$f'(x)=5*3(x^{3-1})$

f'(x)=15x^2

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Example Question #3 : Understanding The Limiting Process.

Differentiate $f(x)=\frac{1}{x^2}$ .

Possible Answers:

2ln(x^2)

$\frac{1}{x^3}$

$\frac{-2}{x}$

ln(x^2)

$\frac{-2}{x^3}$

Correct answer:

$\frac{-2}{x^3}$

Explanation:

Use the product rule:

$f(x)=\frac{1}{x^2}= x^{-2}$

$f'(x)=-2*1(x^{-2-1})= {-2}*{x^{-3}}= \frac{-2}{x^3}$

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AP Calculus AB : Functions, Graphs, and Limits

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #3 : Estimating Limits From Graphs Or Tables

Example Question #4 : Estimating Limits From Graphs Or Tables

Example Question #1 : Understanding The Limiting Process.

Example Question #171 : Functions, Graphs, And Limits

Example Question #2 : Understanding The Limiting Process.

Example Question #1 : Understanding The Limiting Process.

Example Question #1 : Understanding The Limiting Process.

Example Question #171 : Functions, Graphs, And Limits

Example Question #2 : Understanding The Limiting Process.

Example Question #3 : Understanding The Limiting Process.

All AP Calculus AB Resources

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