Question

4TH TIME ASKING THIS!!! Please help me! Someone pleaseeee. I need the correct answers. I don’t want to fail

Answer 1

Answer:

The functions are inverses; f(g(x)) = x ⇒ answer D

$h^{-1}(x)=\sqrt{\frac{x+1}{3}}$ ⇒ answer D

Step-by-step explanation:

* Lets explain how to find the inverse of a function

- Let f(x) = y

- Exchange x and y

- Solve to find the new y

- The new y = $f^{-1}(x)$

* Lets use these steps to solve the problems

∵ $f(x)=\sqrt{x-3}$

∵ f(x) = y

∴ $y=\sqrt{x-3}$

- Exchange x and y

∴ $x=\sqrt{y-3}$

- Square the two sides

∴ x² = y - 3

- Add 3 to both sides

∴ x² + 3 = y

- Change y by $f^{-1}(x)$

∴ $f^{-1}(x)=x^{2}+3$

∵ g(x) = x² + 3

∴ $f^{-1}(x)=g(x)$

∴ The functions are inverses to each other

* Now lets find f(g(x))

- To find f(g(x)) substitute x in f(x) by g(x)

∵ $f(x)=\sqrt{x-3}$

∵ g(x) = x² + 3

∴ $f(g(x))=\sqrt{(x^{2}+3)-3}=\sqrt{x^{2}+3-3}=\sqrt{x^{2}}=x$

∴ f(g(x)) = x

∴ The functions are inverses; f(g(x)) = x

* Lets find the inverse of h(x)

∵ h(x) = 3x² - 1 where x ≥ 0

- Let h(x) = y

∴ y = 3x² - 1

- Exchange x and y

∴ x = 3y² - 1

- Add 1 to both sides

∴ x + 1 = 3y²

- Divide both sides by 3

∴ $\frac{x + 1}{3}=y^{2}$

- Take √ for both sides

∴ ± $\sqrt{\frac{x+1}{3}}=y$

∵ x ≥ 0

∴ We will chose the positive value of the square root

∴ $\sqrt{\frac{x+1}{3}}=y$

- replace y by $h^{-1}(x)$

∴ $h^{-1}(x)=\sqrt{\frac{x+1}{3}}$