# Thread: Inverse of a a function

1. ## Inverse of a a function

Hi, how do you find the inverse of this function: $f(x) = \frac{15}{2x^2 + 11x +5}$ ?

2. ## Re: Inverse of a a function

Hello, sakonpure6!

How do you find the inverse of this function? . $f(x) = \frac{15}{2x^2 + 11x +5}$

Use the prescribed procedure.
[However, the result will not be a function.]

We have: . $y \;=\;\frac{15}{2x^2+11x+5}$

"Switch" variables: . $x \;=\;\frac{15}{2y^2 + 11y + 5}$

Solve for $y\!:$

. . $x(2y^2+11y + 5) \:=\:15 \quad\Rightarrow\quad 2xy^2 + 11xy + 5x - 15 \:=\:0$

Quadratic Formula: . $y \;=\;\dfrac{-11x \pm\sqrt{(11x)^2-4(2x)(5x-15)}}{2(2x)}$

. . . . . . . . . . . . . . . $y \;=\;\frac{-11x \pm \sqrt{121x^2 - 40x^2 + 120x}}{4x}$

. . . . . . . . . . . . . . . $y \;=\;\frac{-11x \pm\sqrt{81x^2 + 120x}}{4x}$

Therefore: . $f^{-1}(x) \;=\;\frac{-11x \pm \sqrt{81x^2+120x}}{4x}$

3. ## Re: Inverse of a a function

Oh I see.. i got stuck on the part with the quadratic. Thank you!

4. ## Re: Inverse of a a function

Originally Posted by sakonpure6
Hi, how do you find the inverse of this function: $f(x) = \frac{15}{2x^2 + 11x +5}$ ?
Did you plot it? You'll notice that f is not bijective over it's entire domain so in order to establish an inverse you will have to restrict the domain of f to an interval, or union of intervals, where it is.

Once you've done that you can solve the quadratic equation y=f(x) for x in terms of y. This gives you x=g(y). Replace every occurrence of y in g(y) with x to obtain g(x).

$f^{-1}(x)=g(x)$ for x in the restricted domain you chose.