# Inverse

• Dec 11th 2008, 07:46 PM
Pre Lives
Inverse and 2 Word Problems (Algebra 2)
I'm having a hard time with this inverse problem and I'm not sure whether I have completely found the inverse or if I did something wrong.

Find inverse: y= 4x^2 - 10x

This is what I have done so far step by step

x= 4y^2 - 10y
0= 4y^2 - 10y - x
Then set up quadratic formula (used power of 1/2 because i don't know how to put sq root on here)
10 +(-10^2 - 4(4)(-x))^(1/2)/8
10 +(100 + 16x)^(1/2)/8 This is where I got stuck

Edit:
Ran into more trouble with 2 word problems:

1st problem
Write an equation that can be used to solve the problem. The answer the question asked. A group of college students are volunteering for habitat for humanity during their spring break. They are putting the finishing touches on a house they built. Working alone, Kathy can paint a certain room in 4 hours. Rhonda can paint the same room in 6 hours. How long will it take them working together to paint the room?

2nd problem
The sum of the reciprocal of a number and the reciprocal of 6 less than the number is 8 times the reciprocal of the original number. Find the original number.
• Dec 11th 2008, 11:00 PM
earboth
Quote:

Originally Posted by Pre Lives
I'm having a hard time with this inverse problem and I'm not sure whether I have completely found the inverse or if I did something wrong.

Find inverse: y= 4x^2 - 10x

This is what I have done so far step by step

x= 4y^2 - 10y
0= 4y^2 - 10y - x
Then set up quadratic formula (used power of 1/2 because i don't know how to put sq root on here)
10 +(-10^2 - 4(4)(-x))^(1/2)/8
10 +(100 + 16x)^(1/2)/8 This is where I got stuck

...

There only exist an inverse function if the original function doesn't change its monotony. Since your function first is monotonically decreasing and after the vertex it is monotonically increasing you have to split the domain into parts such that the function is monotonic.

1. Calculate the coordinates of the vertex by completing the square:

$y=4x^2-10x +\dfrac{25}4 - \dfrac{25}4 = 4 \left(x-\dfrac54 \right)^2-\dfrac{25}4$
V
Thus the vertex is at $V \left(\frac54\ ,\ -\frac{25}4 \right)$

2. Now you can split your function into two monotonic parts:

$f(x)=\left\{\begin{array}{l} 4 \left(x-\dfrac54 \right)^2-\dfrac{25}4\ ,\ x < \dfrac54\ ,\ f(x) \geq -\dfrac{25}4 \\ 4 \left(x-\dfrac54 \right)^2-\dfrac{25}4\ ,\ x \geq \dfrac54\ ,\ f(x) \geq -\dfrac{25}4 \end{array}\right.$

For each of these parts exist an inverse function.

3. Swap the variables and solve the equation for y:

$x=4y^2-10y~\implies~y = \dfrac54\pm\ \dfrac14 \sqrt{25+4x}$

Swap the domain and range of the original function too that you get the domain and range of the inverse function(s):

4. Finally you have:

$f^{-1}(x)=\left\{\begin{array}{l}\dfrac54+\ \dfrac14 \sqrt{25+4x}\ ,\ x\geq -\dfrac{25}4\ ,\ y \geq \dfrac54 \\ \dfrac54-\ \dfrac14 \sqrt{25+4x}\ ,\ x\geq -\dfrac{25}4\ ,\ y < \dfrac54\end{array}\right.$

5. The graphs of the original function and its inverse functions must be reflections over the line y = x (firste median?). See attachment.