# Thread: Hi I need help on inverse functions and other things

1. ## Hi I need help on inverse functions and other things

Hi my name is Cindy and I am new here and I need help on my math homework.

The first one says which of the following fuctions f(x) = sqrt 19 - x and g(x) = sqrt 25 - x^2 has an inverse. Find a rule for the inverse and explain why the other function has no inverse.

I don't understand inverses it would be nice if you could explain to me how you get the inverse.

Also I need help on this problem : Let f(x) = x^2, g(x) = x - 2 and h(x) = 4x, find the following:

A) f(g(h(x)) B) g(h(f(x))

2. Hello, Cindy!

Welcome aboard!
Here's the first one . . .

Which of the following fuctions has an inverse?
$f(x) \:= \:\sqrt{19 - x}$ and $g(x) \:= \:\sqrt{25 - x^2}$
Find a rule for the inverse and explain why the other function has no inverse.
To find the inverse of a function $f(x)$:
. . (1) Replace $f(x)$ with $y$
. . (2) Interchange the $x's$ and $y's$
. . (3) Solve for $y$
. . (4) The resulting function is $f^{-1}(x)$

We have: . $f(x) \:=\:\sqrt{19-x}$

$(1)\;\;y \;=\;\sqrt{19-x}$

$(2)\;\;x\;=\;\sqrt{19-y}$

$(3)$ .Square both sides: . $x^2 \;=\;19-y\quad\Rightarrow\quad y \;=\;19-x^2$

$(4)\;\text{Therefore: }\;f^{-1}(x) \;=\;19 - x^2$

We have: . $g(x) \:=\:\sqrt{25-x^2}$

$(1)\;\;y \;=\;\sqrt{25-x^2}$

$(2)\;\;x \;=\;\sqrt{25-y^2}$

$(3)\;\text{Square both sides: }\;x^2 \:=\:25-y^2\quad\Rightarrow\quad y^2 \:=\:25-x^2$

Then: . $y \;=\;\pm\sqrt{25-x^2}$ . . . This is not a function!

Each value of $x$ produces two values of $y$.
. . and a function must be single-valued.