# Calculus 1 : Other Writing Equations

## Example Questions

← Previous 1 3 4 5 6

### Example Question #1 : Other Writing Equations

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

(x-1)/3

undefined

(y-1)/3

3x-1

(y+1)/3

(y-1)/3

Explanation:

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

### Example Question #1 : Other Writing Equations

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

line with slope 6
circle with diameter 36
line with slope -6
Explanation:

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.

### Example Question #1 : How To Write Equations

Find the equation tangent to at

Equation of tangent line is undefined

Explanation:

First, find the y coordinate of the function at

Thus, we have the point .

To find the slope of the tangent line, take the derivative and then plug in

Thus, our slope is .

Knowing that the general formula of a straight line is where  is the slope and that slope is . Thus, . To find , plug in the coordinate and then solve for .

So now the equation of the tangent line becomes

### Example Question #3 : How To Write Equations

Take the indefinite integral of

Indefinite integral does not exist

Explanation:

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

then perform the indefinite integral the normal way you would do

### Example Question #191 : Equations

Find the limit of .

Explanation:

In this limit, we have  as .

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

.

If we evaluate at , we don't have  anymore and we get

.

### Example Question #2 : How To Write Equations

Find

Explanation:

Because we have a  in our limit, we can use L'Hospital's Rule.  Applying this, we get

.

Evaluating this as  we get

### Example Question #3 : How To Write Equations

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

y = 13694x – 153003

y = 153103x – 13694

y = 55x + 13382

y = 2848x + 34

y = 13694x – 153003

Explanation:

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

### Example Question #8 : Other Writing Equations

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

Explanation:

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:

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

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.

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:

### Example Question #9 : Other Writing Equations

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

Explanation:

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:

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:

### Example Question #2 : How To Write Equations

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

Explanation:

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,

.

← Previous 1 3 4 5 6