Algebra - How to find inverse variation

varies inversely as the square root of . If , then . Find if (nearest tenth, if applicable).

Explanation

The variation equation is $A = \frac{K}{\sqrt{x}}$ for some constant of variation .

Substitute the numbers from the first scenario to find :

$45= \frac{K}{\sqrt{15}}$

$K = 45 \sqrt{15}$

The equation is now $A = \frac{45 \sqrt{15}}{\sqrt{x}}$ .

If , then

$A = \frac{45 \sqrt{15}}{\sqrt{30}} \approx 31.8$

Find the inverse:

$y= \frac{1}{4}x + \frac{3}{2}$

$y= -\frac{1}{4}x + \frac{3}{2}$

$y= \frac{1}{4}x- \frac{3}{2}$

Explanation

To find the inverse, first interchange the x and y variables in the equation.

Solve for y. Add 6 on both sides.

Simplify.

Divide by four on both sides.

$\frac{x+6}{4}=\frac{4y}{4}$

Simplify both sides.

$y=\frac{x+6}{4} = \frac{1}{4}x + \frac{6}{4}= \frac{1}{4}x + \frac{3}{2}$

The inverse is: $y= \frac{1}{4}x + \frac{3}{2}$

Find the inverse of the following algebraic function:

$f(x)=\frac{1}{x-3}$

$y=\frac{1}{x}+3$

$y=\frac{1}{x}-3$

$y=x+\frac{1}{3}$

$y=x-\frac{1}{3}$

Explanation

To find the inverse, switch the placement of the and variables:

$f(x)=\frac{1}{x-3}$

$y=\frac{1}{x-3}$

$x=\frac{1}{y-3}$

Next, should be isolated, providing the inverse function:

$\frac{y-3}{1}\cdot x=\frac{1}{y-3}\cdot \frac{y-3}{1}$

$\frac{xy}{x}=\frac{1+3x}{x}$

$y=\frac{1+3x}{x}$

$y=\frac{1}{x}+3$

is a one-to-one function specified in terms of a set of $\left ( x,y \right )$ coordinates:

A = $\left { \left ( 0,0 \right ), \left ( 1,2\right ),\left ( 2,4 \right ), \left ( 3,6 \right )\right }$

Which one of the following represents the inverse of the function specified by set A?

B = $\left { \left ( 0,0 \right ), \left ( 0,1 \right ), \left ( 0,2 \right ), \left ( 0,3 \right ) \right }$

C = $\left { \left ( 0,0 \right ), \left ( 2,1 \right ), \left ( 4,2 \right ), \left ( 6,3 \right ) \right }$

D = $\left { \left ( 1,1 \right ), \left ( 2,1 \right ), \left ( -2,1 \right ), \left ( 3,1 \right ) \right }$

E = $\left { \left ( -2,1 \right ), \left ( -2,5 \right ), \left ( -2,3 \right ), \left ( -2, -2 \right ) \right }$

F = $\left { \left ( 2,0 \right ), \left ( 2,-3 \right ),\left ( 2,5 \right ), \left ( 2,-7 \right ) \right }$

Set C

Set B

Set D

Set E

Set F

Explanation

The set A is an one-to-one function of the form

One can find $f^{-1}\left ( x \right )$ by interchanging the and coordinates in set A resulting in set C.

The amount of lonliness you feel varies inversely with the number of friends you have. If having 4 friends gives you a 10 on the lonliness scale, how much lonliness do you feel if you have 100 friends?

$\small 0.4$

$\small 4$

$\small 2.5$

$\small 25$

Explanation

If lonliness and friends are inversely proportional, we can set up an equation to solve for some missing constant, k. To make things easier to write, let's use the variable L for the lonliness scale, and the variable F for how many friends you have. First we can just set up the equation:

$\small L = \frac{k}{F}$ . We know that having 4 friends gives you a 10 on the lonliness scale, so we can plug those values in to start solving for k:

$\small 10 = \frac{k}{4}$ to solve, multiply both sides by 4:

$\small 40 = k$ Now that we know the constant is 40, we can figure out how much lonliness, L, corresponds to having 100 friends by setting up a new formula. We are still generally using $\small L = \frac{k}{F}$ , and we know k is 40 and F is 100: