# Thread: Final Exam Study Guide Help

1. ## Final Exam Study Guide Help

I am working on my final exam study guide for college algebra and have a few I need help with.

1 "Fill in the blank. Given that f(x) = -4x-2 and g(x)=x^2=9. (fog)(-9)= ?

2 "Find the inverse of the funtion f(x)= 3x-5/6. Then Fill in the blank below. F^-1(x)= ?

3 The intensity of illumination from a light source varies inversley with square of the distance from the source. Suppose that a reading lamp has an intensity of 5000 lumens at a distance of 4 feet.

The intesnity of the lamp is how many lumens at a distance of 2 feet?

The intensity of the lame is 300 lumens at a distnace of how many feet?

2. Originally Posted by Ty Durdan
I am working on my final exam study guide for college algebra and have a few I need help with.

1 "Fill in the blank. Given that f(x) = -4x-2 and g(x)=x^2=9. (fog)(-9)= ?
You shoud know that:

$(f \circ g)(x)=f(g(x))=f(x^2)=-4x^2-2$

RonL

3. [quote=Ty Durdan;64733]2 "Find the inverse of the funtion f(x)= 3x-5/6. Then Fill in the blank below. F^-1(x)= ?

If $y=f(x)=3x-5/6$, then $x=y/3+5/18$, so:

[tex]f^-1(y)=y/3+5/18[tex]

or rewriting using $x$ as the variable:

$f^-1(x)=x/3+5/18$

RonL

4. Originally Posted by Ty Durdan
3 The intensity of illumination from a light source varies inversley with square of the distance from the source. Suppose that a reading lamp has an intensity of 5000 lumens at a distance of 4 feet.

The intesnity of the lamp is how many lumens at a distance of 2 feet?

The intensity of the lame is 300 lumens at a distnace of how many feet?
Let I = intensity and D = distance, then

$I \propto \frac{1}{D^2}$

=> $I=\frac{k}{D^2}$

When I = 5000, D = 4, and we solve for k:

$5000=\frac{k}{4^2}$

$k=80000$

=> $I=\frac{80000}{D^2}$

For the first question, D = 2:
$I=\frac{80000}{2^2}$

$I=20000$ lumens

For the second, I = 300:
$300=\frac{80000}{D^2}$

$D^2=\frac{80000}{300}$

$D=\frac{20 \sqrt{6}}{3}$ feet (we obviously only need a positive solution)