# College Algebra Questions?

A stone is dropped into a pond, causing a circular ripple that is expanding at a rate of 10 ft/sec. Describe the area of the circle, A , as a function of time, t .

Enter an exact answer in terms of pie.

Check whether f and g are inverses of each other.

f(x) = 2x + 2 and x-2/2

(f*g)(x)

(g*f)(x)

Find the inverse for the given function.

5/x+7

Relevance

Hi Mandy,

(1) The key thing you need to know in this is that when the ripple is expanding, that it is the radius of the circle that is increasing. Thus, as a function of time, the radius, r(t) = 10t, assuming we're starting at a radius = 0 at t=0.

Now just apply the area formula:

A = pi r^2 = pi * (10t)^2 = pi * 100 t^2 = 100pi t^2.

(2) For your second problem, to determine whether functions are inverses of each other you have to show that f*g = g*f = x. That is, no matter which function you start with, by composing it with the other, you back the original input.

In your case, you f(x) = 2x + 2 and g(x) = (x-2)/2 [I'll assume you meant to group the x-2 in the g(x), as it would be kind of silly to start off with an expression that is so easily reduceable.]

f*g(x) = f(g(x)) = f([(x-2)/2]) = 2*[(x-2)/2] + 2 = (x-2) + 2 = x. CHECK!

g*f(x) = g(f(x)) = g(2x+2) = [(2x+2)-2]/2 = 2x/2 = x. CHECK! Thus, f and g are inverses.

(3) To find the inverse of a given function in x, simply set it equal to y. Then swap x and y, and solve for y:

(a) In case you intended to write the problem just as you typed it: y = 5/x + 7; then the inverse satisfies the relation: x = 5/y + 7. Now make it explicit in terms of y:

x = 5/y + 7;

x - 7 = 5/y

y(x-7) = 5

y = 5/(x-7).

(b) Just in case you meant to type: y = 5/(x+7); then the inverse satisfies the relation: x = 5/(y+7). Now make it explicit in terms of y:

x = 5/(y+7);

x(y+7) = 5;

y + 7 = 5/x

y = 5/x - 7.

Both (a) and (b) can be checked by using the method in (2) above.

To help reinforce your understanding of inverse functions, I've searched and found a webpage and a video tutorial that address problems similar to this one, and I thought they might be helpful to you. I've listed them below.

As always, if you need more help, please clarify where you are in the process and what's giving you trouble. I'd be more than happy to continue to assist.

• First Part:

Let A(t) be the area function of the circle defined by a circular ripple, whose size (diameter) is expanding at 10 ft/sec. Then:

A(t) = 25πt²

Second Part:

f(x) = 2x + 2

g(x) = (x-2)/2

(f o g)(x) = 2([x-2]/2) + 2 = x

(g o f)(x) = ([2x + 2] - 2)/2 = x

Check: f & g are inverses of each other

Third Part:

y = 5/x + 7

y - 7 = 5/x

x = 5/(y - 7)

For f(x) = 5/x + 7

g(x) = 5/(x - 7) is an inverse

• If the radius is r, then dr/dt = 10

Integrate that to get r = 10t + C

Presumably, r = 0 when t = 0, so c = 0, therefore r = 10t

A = pi * r^2

A = pi * (10t)^2

A = pi * 100t^2

y = 2x + 2

y - 2 = 2x

x = (y - 2) / 2

The inverse is:

y = (x - 2) / 2

Note that x - 2/2 is the same as x - 1, so the answer is NO, that's not the inverse.

But if you add brackets to make it (x - 2) / 2, the answer is YES, that's the inverse.

y = (5/x) + 7

y - 7 = 5/x

x(y - 7) = 5

x = 5 / (y - 7)

The inverse is:

y = 5 / (x - 7)

Or did you mean:

y = 5 / (x + 7)

y(x + 7) = 5

xy + 7y = 5

xy = 5 - 7y

x = (5 - 7y) / y

The inverse is:

y = (5 - 7x) / x

y = (5/x) - 7

Or did you mean:

• Anonymous
8 years ago

is the increase the circumference or the radius or evn possibly the area?

what is the relation between R and A ?