Integrals - AP Calculus AB

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Question

A ball is thrown into the air. It's height, after t seconds is modeled by the formula:

h(t)=-15t^2+30t feet.

At what time will the velocity equal zero?

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Answer

In order to find where the velocity is equal to zero, take the derivative of the function and set it equal to zero.

h(t) = –15t2 + 30t

h'(t) = –30t + 30

0 = –30t + 30

Then solve for "t".

–30 = –30t

t = 1

The velocity will be 0 at 1 second.

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Question

Write the domain of the function.

$f(x)=\frac{(x^{4}-81)^{1/2}}{x-4}$

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Answer

The answer is ${x\mid x\neq 4\ and\ \left | x \right |\geq 3}$

The denominator must not equal zero and anything under a radical must be a nonnegative number.

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Question

Find $\lim _{x\rightarrow 0}f(x)$

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Answer

The one side limits are not equal: left is 0 and right is 3

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Question

Which of the following is a vertical asymptote?

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Answer

When approaches 3, approaches $\pm \infty$ .

Vertical asymptotes occur at values. The horizontal asymptote occurs at

.

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Question

Evaluate the following limit:

$\lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}$

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Answer

First, let's multiply the numerator and denominator of the fraction in the limit by $\frac{1}{x^{4}}$ .

$\lim_{x\rightarrow \infty }\frac{1-x^4}{x^2-4x^4}$ $=\lim_{x\rightarrow \infty }\frac{(\frac{1}{x^4})(1-x^4)}{\frac{1}{x^{4}}(x^2-4x^4)}$

$=\lim_{x\rightarrow \infty }\frac{\frac{1}{x^4}-1}{\frac{1}{x^2}-4}$

As becomes increasingly large the $\frac{1}{x^{4}}$ and $\frac{1}{x^{2}} ^{}$ terms will tend to zero. This leaves us with the limit of $\frac{1}{4}$ .

$\lim_{x\rightarrow \infty }\frac{-1}{-4}=\frac{1}{4}$ .

The answer is $\frac{1}{4}$ .

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Question

Let and be inverse functions, and let

$\f(3)=4 \g(3)=-5 \g(4)=3 \f'(3)=-2 \f'(4)=4 \g'(3)=-1$ .

What is the value of ?

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Answer

Since and are inverse functions, . We can differentiate both sides of the equation with respect to to obtain the following:

$g'(f(x))\cdot f'(x)=1$

We are asked to find , which means that we will need to find such that . The given information tells us that , which means that . Thus, we will substitute 3 into the equation.

$g'(f(3))\cdot f'(3)=1$

The given information tells us that.

The equation then becomes $g'(4)\cdot (-2)=1$ .

We can now solve for .

$g'(4)=-\frac{1}{2}$ .

The answer is $-\frac{1}{2}$ .

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Question

Using the fundamental theorem of calculus, find the integral of the function from to .

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Answer

The fundamental theorem of calculus is, $\int_{a}^{b} f(x)\ dx=F(b)-F(a)$ , now lets apply this to our situation.

$\int_{0}^{1} x^3+2x+9\ dx$

We can use the inverse power rule to solve the integral, which is $\int x^n\ dx =\frac{x^{n+1}}{n+1}+C$ .

$\int_{0}^{1} x^3+2x+9\ dx=\frac{x^4}{4}+x^2+9x \Big|^{1}_{0}$

$=\left[ \frac{1^4}{4}+1^2+9(1)\right]-\left[\frac{0^4}{4}+0^2+9(0) \right ]$

$=\frac{1}{4}+1+9-0=\frac{41}{4}$

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Question

$f(x) = \frac{x+1}{\sqrt{x^2 -4}}$

What are the horizontal asymptotes of ?

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Answer

Compute the limits of as approaches infinity.

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Question

What is the value of the derivative of $f(x)=x^3+6x^2+\frac{3}{4}x$ at x=1?

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Answer

First, find the derivative of the function, which is:

$f{}'(x)=3x^2+12x+\frac{3}{4}$

Then, plug in 1 for x:

$3(1^2)+12(1)+\frac{3}{4}$

The result is $\frac{63}{4}$ .

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Question

Write the equation of a tangent line to the given function at the point.

y = ln(x2) at (e, 3)

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Answer

To solve this, first find the derivative of the function (otherwise known as the slope).

y = ln(x2)

y' = (2x/(x2))

Then, to find the slope in respect to the given points (e, 3), plug in e.

y' = (2e)/(e2)

Simplify.

y'=(2/e)

The question asks to find the tangent line to the function at (e, 3), so use the point-slope formula and the points (e, 3).

y – 3 = (2/e)(x – e)

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Question

If $f(x)=4, f'(x)=3, g(x)=2, \ and\ g'(x)=1$ then find .

$h(x)=\frac{2f}{g}$

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Answer

The answer is 1.

$h(x)=\frac{2f}{g}$

$h'(x)=\frac{2(f'(x)g(x)-f(x)g'(x))}{g^{2}}$

$h'(x)=\frac{2(32-41)}{4} = 1$

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Question

If $f(x)=1, f'(x)=2, g(x)=3, \ and\ g'(x)=4$ then find .

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Answer

The answer is 10.

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Question

Find the equation of the tangent line at on graph

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Answer

The answer is

$f(x)=x^{2}+2x-8$

$f'(x)=2x+2$

(This is the slope. Now use the point-slope formula)

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Question

Find the equation of the tangent line at (1,1) in

$f(x)=ax^{2}+bx+c$

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Answer

The answer is

$f(x)=ax^{2}+bx+c$

$f'(x)=2ax+b$

$f'(1)=2a(1)+b = 2a+b$

(This is the slope. Now use the point-slope formula.)

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Question

If $f(x)=\cos x$ then

$f^{}\left ( \frac{\pi }{2} \right )=$

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Answer

The answer is .

We know that

$\frac{\pi }{2}=90^{\circ}$

so,

$f'\left ( \frac{\pi }{2}\right )=-sin(90^{\circ})=-1$

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Question

Differentiate $y=3\csc x+5\sin x$

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Answer

The answer is $-3\csc x\cot x+5\cos x$

We simply differentiate by parts, remembering our trig rules.

$y=3\csc x+5\sin x$

$y^{}= -3\csc x\cot x+5\cos x$

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Question

Find the equation of the tangent line at when

$y=\frac{x^{3}-2x(x^{1/2})}{x}$

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Answer

The answer is

$y=\frac{x^{3}-2x(x^{1/2})}{x}$ let's go ahead and cancel out the 's. This will simplify things.

$y=x^{2}-2(x^{1/2})$

$y'=2x-\frac{1}{(x^{1/2})}$

$y'(1)=2(1)-\frac{1}{(1^{1/2})} =1$ this is the slope so let's use the point slope formula.

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Question

Differentiate $y=\frac{t+2}{t^{2}-4t-12}$

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Answer

We see the answer is $y'=\frac{-1}{(t-6)^{2}}$ after we simplify and use the quotient rule.

$y=\frac{t+2}{t^{2}-4t-12}$ we could use the quotient rule immediatly but it is easier if we simplify first.

$y=\frac{t+2}{(t+2)(t-6)}$

$y=\frac{1}{(t-6)}$

$y=(t-6)^{-1}$

$y'=-(t-6)^{-2}$

$y'=\frac{-1}{(t-6)^{2}}$

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Question

Find $\lim_{x\rightarrow \infty }\frac{-2x^3+x^2-2}{x^2+10}$

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Answer

When taking limits to infinity, we usually only consider the highest exponents. In this case, the numerator has $-2x^3$ and the denominator has $x^2$ . Therefore, by cancellation, it becomes $-2x$ as approaches infinity. So the answer is $-\infty$ .

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Question

What is the first derivative of the function $\large h(x)=x^{\frac{1}{x}}$ ?

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Answer

First, let .

$y=x^{\frac{1}{x}}$

We will take the natural logarithm of both sides in order to simplify the exponential expression on the right.

$\ln y=\ln x^{\frac{1}{x}}$

Next, apply the property of logarithms which states that, in general, $\log x^a=a\log x$ , where is a constant.

$\ln y = \frac{1}{x}\ln x$

We can differentiate both sides with respect to .

$\frac{d}{dx}\[ln y]=\frac{d}{dx}[\frac{1}{x}\ln x]$

We will need to apply the Chain Rule on the left side and the Product Rule on the right side.

$\frac{d}{dy}[\ln y]\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{d}{dx}[\ln x]+\ln x\cdot \frac{d}{dx}[\frac{1}{x}]$

$\frac{1}{y}\cdot \frac{dy}{dx}=\frac{1}{x}\cdot \frac{1}{x} + \ln x\cdot \frac{-1}{x^{2}}$

Because we are looking for the derivative, we must solve for $\frac{\mathrm{d}y }{\mathrm{d} x}$ .

$\frac{dy}{dx}=y\cdot \frac{1}{x^{2}}(1-\ln x)$

However, we want our answer to be in terms of only. We now substitute $x^{\frac{1}{x}}$ in place of .

$\frac{dy}{dx}=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$

Since we let , we can replace $\frac{\mathrm{d} y}{\mathrm{d} x}$ with .

The answer is $h'(x)=\frac{x^{\frac{1}{x}}}{x^{2}}(1-\ln x)$ .

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