inverse of function

Question

Thanks for the help! Here is the solution..

i had a problem: $$f(x)=\frac{(\sqrt x+8)}{(5-\sqrt x)}$$

i had to find the inverse, so lets begin...

1) first i write in terms of $y$

$$y=\frac{(\sqrt x+8)}{(5-\sqrt x)}$$

2) now try to get $x$ by itself

$$(5-\sqrt x)y=\sqrt x+8$$

3)Distribute y along the left hand side both sides

$$(5y-y\sqrt x)=\sqrt x+8$$

4)subtract 8 and from both sides and add $$ y\sqrt x $$ to both sides

$$ 5y-8=y\sqrt{x} + \sqrt x$$

5) factor right hand side

$$ 5y-8=\sqrt{x} (y + 1)$$

6)divide out $$ (y+1) $$ from right side

$$ ((5y-8)/(y+1))=\sqrt x $$

7) square both sides leaving $$ x=((5y-8)/(y+1))^2$$

8) so the solution is $$f^-1(x)=((5x-8)/(x+1))^2$$

Answer 1

I have a problem:

$$f(x)= \frac{\sqrt{x}+8}{5-\sqrt{x}}$$

I have to find the inverse but my calculations are off. I have listed what I did below, could someone tell me where I went wrong? Thank You in advance.

1) Write in terms of y

$$y = \frac{\sqrt{x}+8}{5-\sqrt{x}}$$

2) Now try to get $x$ by itself

$$ y(5-\sqrt{x}) = \sqrt{x}+8 $$

3) Multiply the $y$ through the expression

$$ 5y-y\sqrt{x} = \sqrt{x}+8 $$

5) Subtract 8 from both sides

$$ 5y-y\sqrt{x} -8= \sqrt{x} $$

5) Add $y\sqrt{x}$ to both sides

$$ 5y - 8= \sqrt{x} +y\sqrt{x}$$

6) Factor out $\sqrt{x}$

$$ 5y - 8= \sqrt{x}(1+y)$$

7) Divide by $(1+y)$

$$ \frac{5y - 8}{(1+y)}= \sqrt{x}$$

8) Square both sides

$$ \frac{(5y - 8)^2}{(1+y)^2}= x$$

9) Replace $x$ with $y$

$$ f^{-1} = \frac{(5x - 8)^2}{(1+x)^2}$$