How to write equations - AP Calculus AB

Card 0 of 784

Question

Given the one-to-one equation f(x)=3x+1, the inverse function f-1(y)=

Answer

Before we solve the problem by computation, let's look at the answer choices and see if we can eliminate any answer choices. We know that an inverse function must exist because f(x) is one-to-one, so we can eliminate the answer choice "undefined." Next, we know that the inverse function has to be in terms of y, so we can eliminate the two answer choices with an "x."

Now we can look at the two remaining answer choices. Let y=f(x) and solve for x to find the inverse.

So f(x)=y=3x+1. Solve for x.

x=(y-1)/3

Therefore our answer is (y-1)/3.

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Question

Describe the graph of the polar equation r=-6.

Answer

The polar equation formula here is x2+y2=r2.

This is the formula for a cirlce, so we can eliminate the two "line" answers.

Plugging in r=-6 into the equation gives x2+y2=36, which describes the graph of a circle with radius 6.

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Question

What is the equation of the tangent line at x = 15 for f(x) = x4 + 5x2 + 44x – 3?

Answer

First we must solve for the general derivative of f(x) = f(x) = x4 + 5x2 + 44x – 3.

f'(x) = 4x3 + 10x + 44

Now, the slope of the tangent line for f(15) is equal to f'(15):

f'(15) = 4(15)3 + 10 15 + 44 = 13694.

To find the tangent line, we need at least one point on the line. To find this, we can use f(15) to get the y value of the point of tangency, which will suffice for our use:

f(15) = 154 + 5(15)2 + 44 15 – 3 = 50625 + 1125 + 660 – 3 = 52407

Now, using the point-slope form of the line, we get:

y - 52407 = 13694 * (x – 15)

Simplify:

y – 52407 = 13694x – 205410

y = 13694x – 153003

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Question

Find the equation tangent to $y=x\sin x$ at $x=\pi$

Answer

First, find the y coordinate of the function at $x=\pi$

$y=\pi \sin \pi = 0$

Thus, we have the point $(\pi,0)$ .

To find the slope of the tangent line, take the derivative and then plug in $x=\pi$

${y}'=\sin x +x\cos x$

${y}'(\pi)=\sin \pi +\pi\cos \pi=-\pi$

Thus, our slope is $-\pi$ .

Knowing that the general formula of a straight line is $y=mx+b$ where is the slope and that slope is $-\pi$ . Thus, $y=-\pi x+b$ . To find , plug in the coordinate $(\pi,0)$ and then solve for .

$0=-\pi (\pi)+b$

$b=\pi^2$

So now the equation of the tangent line becomes $y=-\pi x +\pi^2$

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Question

Take the indefinite integral of

$\int x(x^6+7)^2dx$

Answer

First, foil the integral so it is easier to manage.

$\int (x^{13}+14x^7+49x) dx$

then perform the indefinite integral the normal way you would do

$\frac{1}{14}x^{14}+\frac{7}{4}x^8+\frac{49}{2}x^2+C$

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Question

Find the limit of $lim_{x \to 0}\frac{sin(2x)}{x}$ .

Answer

In this limit, we have $\frac{0}{0}$ as $x\rightarrow 0$ .

Because of this, we can use L'Hospital's rule. Differentiating the top and the bottom of the function, we get

$\frac{2cos(2x)}{1}$ .

If we evaluate at , we don't have $\frac{0}{0}$ anymore and we get

$\frac{2cos(0)}{1}=\frac{2*1}{1}=2$ .

So the answer is 2.

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Question

Find $lim_{x \to 0}\frac{3^x-2^x}{x}.$

Answer

Because we have a $\frac{0}{0}$ in our limit, we can use L'Hospital's Rule. Applying this, we get

$lim _{x \to 0} \frac{3^x-2^x}{x}=lim_{x \to 0}\frac{ln(3)3^x-ln(2)2^x}{1}$ .

Evaluating this as ${x \to 0}$ we get

$\frac{ln(3)3^0-ln(2)2^0}{1}=ln(3)-ln(2)=ln(3/2).$

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Question

For the following function, use implicit differentiation to find the equation of the tangent line at :

$y^2e^{2x}=3y+x^2$

Answer

Our first step is to apply implicit differentiation to the function to find y’. This derivative tells us the slope of our function and will therefore give us the slope of our tangent line at the given point:

$2yy'e^{2x}+2y^2e^{2x}=3y'+2x$

Rearranging the equation, factoring out y’, and dividing, we find y’ is:

$y'=\frac{2x-2y^2e^{2x}}{2ye^{2x}-3}$

Before we can plug in the values of our given point to find the slope of the tangent line, we must first find its y value by plugging x=0 into the original function, which gives us:

So now that we know our given point is (0,3), we plug these values into our equation for y’, which gives us the slope of the tangent line at that point.

$m=y'(0,3)=\frac{2(0)-2(3)^2e^0}{2(3)e^0-3}=-\frac{18}{3}=-6$

Now that we have our slope (m), an x coordinate (x1), and a y coordinate (y1), we can simply plug these values into the point-slope formula for a line:

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Question

Find two positive numbers whose sum is and whose product is a maximum.

Answer

This problem deals with the concept of optimization. We start by simply writing out the equations described in words by the problem:

The problem asks that we maximize the product of the two unknown numbers, C, but we must first solve our first equation for y. Then we substitute it into our second equation to ensure that it is in terms of x alone. We then take the derivative of the equation with respect to x, and set it equal to 0 to solve for the x value that corresponds to the critical point of the function:

$x+y=450\rightarrow y=450-x$

$450-2x=0\rightarrow x=225$

Because we only know that this x value corresponds to a critical point, and not necessarily whether that point is a minimum or a maximum, we must check the value of the second derivative at this point to see whether the function is concave up or concave down:

Because the value of the second derivative at this point is negative, we know that the function is concave down, so our one and only critical point in this case occurs at a global maximum, verifying that the product of the two numbers is maximized when x=225. Now that we know the value of x, we substitute it into the equation for y to find our second unknown number:

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Question

What is the equation of a line with slope of and -intercept of ?

Answer

To write this equation we need to read the question carefully. Since the slope is we know the slope is represented by .

Also since the y-intercept is , when is then .

Therefore plugging our values into the slope intercept form:

we get,

.

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Question

My friend has dollars at time days. She earns dollars every day, and spends dollars every day (so at day , she has dollars). Write an equation for how much money she has at time .

Answer

My friend starts off with dollars, and every day, she makes a net dollar. This is because she earns three dollars and spends one dollar.

So, we need an equation that reflects that she has *(number of days), which is given by

.

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Question

Identify the inner and the outer functions of the following equation (let be the outer equation and be the inner equation:

$y=\sqrt[3]{1-4x}$

Answer

The first section of the equation to be resolved (in this case 1-4x) is the innermost function and the second section to solve is the outer function. So:

$f(x)=\sqrt[3]{x}$

becuase g(x) must be performed first before plugging it into f(x).

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Question

Identify the inner and the outer functions of the following equation (let be the outer equation and be the inner equation:

Answer

The first section of the equation to be resolved (in this case 2x^3+5) is the innermost function and the second section to solve is the outer function.

So:

becuase g(x) must be performed first before plugging it into f(x).

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Question

Identify the inner and the outer functions of the following equation (let be the outer equation and be the inner equation:

$y=\tan\pi x$

Answer

The first section of the equation to be resolved (in this case pi * x) is the innermost function and the second section to solve is the outer function.

So:

$f(x)=\tan(x)$

$g(x)=\pi x$

becuase g(x) must be performed first before plugging it into f(x).

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Question

Derive the equation for the function of the .

Answer

$\cot(x)=\frac{\cos (x)}{\sin (x)}$

By the quotient rule, the derivative of

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

The derivative of sin is cos and the derivative of cos is -sin. Thus the derivative of cot(x) is

$\frac{-\sin (x)^2-\cos (x)^2}{\sin (x)^2}$

$=-(1+\cot (x)^2)$

By the trigonometric identities this is equal to $-\csc(x)^{2}$ .

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Question

Suppose Charlie deposits per month into an account that contains a starting balance of .

Which of the following is a differential equation that best models the amount of money, , in the account after years?

Answer

Here we are writing a differential equation in the units of dollars per year. This means that we need to reconcile our units to get the amount deposited each year. ,

so the yearly rate of change of money in the account,

$\frac{dM}{dt}=180$ .

The starting balance is our initial condition and does not tell us about the change in money in the account, so is not included in our differential equation.

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Question

Suppose Charlie deposits per month into an account that contains a starting balance of .

Which of the following is a correct initial condition that when coupled with the differential equation from the previous question, will yield a specific solution to the scenario described above?

Answer

Here we are looking for an initial condition that describes the balance of the account at a specific time. We are given the information that the account starts with initially. So at time years, we know that the amount of money in the account is . Therefore, we can write

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Question

Find the implicit derivative $\frac{\mathrm{d} y}{\mathrm{d} x}$ of at the point .

Answer

Use implicit differentiation to find $\frac{\mathrm{d} y}{\mathrm{d} x}$ of which means to take the derivative of each term in the function with its respective part.

It is simpfied to

$4xy+2x^2\frac{\mathrm{d} y}{\mathrm{d} x} + y+x\frac{\mathrm{d} y}{\mathrm{d} x} - 3y^2\frac{\mathrm{d} y}{\mathrm{d} x}=2 -\frac{\mathrm{d} y}{\mathrm{d} x}$ .

It is then further simplified to

$2x^2\frac{\mathrm{d} y}{\mathrm{d} x}+x\frac{\mathrm{d} y}{\mathrm{d} x}-3y^2\frac{\mathrm{d} y}{\mathrm{d} x}+\frac{\mathrm{d} y}{\mathrm{d} x}=2-4xy-y$ ,

then to

$\frac{\mathrm{d} y}{\mathrm{d} x}(2x^2+x-3y^2+1)=2-4xy-y$ ,

then to

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2-4xy-y}{2x^2+x-3y^2+1}$ .

Plugging in into the equation gives the value of $-\frac{1}{2}$ .

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Question

If and , then what is ?

Answer

First we need to find , which is .

Then we find , which equals to

$f''(x^2)=\frac{\mathrm{d} (6xg(x^2))}{\mathrm{d} x}=6g(x^2)+6xf(x^2)2x=6g(x^2)+12x^2f(x^2)$ .

We used the power rule and the product rule to find the derivatives.

Power Rule: $x^n \rightarrow nx^{n-1}$

Product Rule: $f(x)g(x)\rightarrow f(x)g'(x)+g(x)f'(x)$

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Question

$y=x^{x^3+2x^2}$ , find $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

Answer

Our starting function is $y=x^{x^3+2x^2}$ in order to create an easier function we will want to take the natural log of both sides. Taking the natural log will bring the exponet infront creating a product for which we can more easily differentiate.

Next take the natural log of both sides which results in the following,

$\ln(y)=\ln (x)^{x^3+2x^2}=(x^3+2x^2)\ln (x)$ .

Now differentiate both sides with respect to x.

$\frac{1}{y}\frac{\mathrm{d} y}{\mathrm{d} x}=(3x^2+4x)\ln (x)+\frac{1}{x}(x^3+2x^2)=(3x^2+4x)\ln (x)+(x^2+2x)$ , and when one isolates $\frac{\mathrm{d} y}{\mathrm{d} x}$ .

After isolating $\frac{\mathrm{d} y}{\mathrm{d} x}$ make sure to replace the y term in terms of x.

$\ \frac{\mathrm{d} y}{\mathrm{d} x}=[(3x^2+4x)\ln (x)+(x^2+2x)]y\ \=[(3x^2+4x)\ln (x)+(x^2+2x)](x^{x^3+2x^2})$

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