1. ## functions ...

f(x) = √(2x² + 12)

find inverse.

what i was taught was to write f(x) = y

so y = √(2x² + 12)

then switch x and y

so x = √(2y² + 12)

then rearrange for y

so i get

y = (1/2)√(x² - 12)

could anyone confirm this is right or wrong?

2. Originally Posted by djmccabie
f(x) = √(2x² + 12)

find inverse.

what i was taught was to write f(x) = y

so y = √(2x² + 12)
You're right up to here for sure.

then switch x and y

so x = √(2y² + 12)

then rearrange for y

so i get

y = (1/2)√(x² - 12)

could anyone confirm this is right or wrong?
What I was taught after putting f(x) = y is make x the subject, written in terms of y:

$y = \sqrt{2x^2+12}$

$2x^2+12 = y^2$

$
2x^2 = y^2-12$

$
x^2 = \frac{1}{2}(y^2-12)$

$x = \pm \sqrt{\frac{1}{2}(y^2-12)} = \pm \frac{\sqrt2}{2}{\sqrt{y^2-12}}$

Finally replace x with $f^{-1}(x)$ and y with x:

$f^{-1}(x) = \pm \frac{\sqrt2}{2}{\sqrt{x^2-12}}$

Note that the plus/minus sign means it's not one to one and so not a function unless the domain is restricted

3. Originally Posted by djmccabie
f(x) = √(2x² + 12)

find inverse.

what i was taught was to write f(x) = y

so y = √(2x² + 12)

then switch x and y

so x = √(2y² + 12)

then rearrange for y

so i get

y = (1/2)√(x² - 12)

could anyone confirm this is right or wrong?
It should be $y = \pm \sqrt{\frac{x^2 - 12}{2}}$.

4. Originally Posted by e^(i*pi)
You're right up to here for sure.

What I was taught after putting f(x) = y is make x the subject, written in terms of y:

$y = \sqrt{2x^2+12}$

$2x^2+12 = y^2$

$
2x^2 = y^2-12$

$
x^2 = \frac{1}{2}(y^2-12)$

$x = \pm \sqrt{\frac{1}{2}(y^2-12)} = \pm \frac{\sqrt2}{2}{\sqrt{y^2-12}}$

Finally replace x with $f^{-1}(x)$ and y with x:

$f^{-1}(x) = \pm \frac{\sqrt2}{2}{\sqrt{x^2-12}}$

Note that the plus/minus sign means it's not one to one and so not a function unless the domain is restricted

This is actually the same method except you switch x and y at the end