Find the inverse of $$h(x)=8sqrt{8x-3}-1$$ where
$$x ge \frac{3}{8}$$

Question

Find the inverse of $$h(x)=8sqrt{8x-3}-1$$ where 
$$x ge \frac{3}{8}$$

anonymous · Answer

$$h(x)=8 \sqrt{8x-3}-1$$
where $$x \ge \frac{3}{8}$$

and @saifoo.khan I know what you're thinking... I post a question just like this... but I just want to make sure am doing the right thing... :D.

anonymous · Answer

And... I have not been given one with certain conditions in class...

saifoo.khan · Answer

I'm not sure i can solve this one. sorry.

saifoo.khan · Answer

But i will try.

saifoo.khan · Answer

(y+1)/8 = sqrt(8x-3)

take square on both sides..
$$(\frac{y+1}{8})^2 = 8x-3$$

saifoo.khan · Answer

@dpaInc , HELP!

anonymous · Answer

I got...

$$f^{-1}(x)=\frac{1}{8}(\frac{x+1}{2})^2+\frac{3}{8}$$
but I don't know if I have to clean it up...

saifoo.khan · Answer

it's 8 sqrt or 2 sqrt ?

anonymous · Answer

8

saifoo.khan · Answer

Then how you got (x+1)/2 ?

anonymous · Answer

Oh that is suppose to be an 8... Idk why am making up numbers...

saifoo.khan · Answer

Then following up,
$$\frac{(y+1)^2}{64} +3 = 8x$$

saifoo.khan · Answer

brb 10mins

anonymous · Answer

Seems as if you didn't exchange x and y...

apoorvk · Answer

Hmm, where are we stuck now?

anonymous · Answer

I got:
$$\frac{y^{2} + 2x + 193}{512}$$
however, I'm not sure what the purpose of the x ≥ 3/8 part is about...

apoorvk · Answer

well yeah, x $\ge$ 3/8 since restrictions apply for the original equation for the part under the root. otherwise you're right @Calcmathlete , just need to interchange 'y' and 'x' now.

anonymous · Answer

Oh alright.
$$f(x)^{-1} = \frac{x^{2} + 2x + 193}{512}$$

apoorvk · Answer

Almost! just the wee tiny winy thing! :p
 
$\large f^{-1}(x) = \frac{x^{2} + 2x + 193}{512}$

anonymous · Answer

lol. Slight mix up :)

apoorvk · Answer

no worries! :D

anonymous · Answer

Oh thanks guys! I got the same thing :)

anonymous · Answer

np :)

apoorvk · Answer

Yo!