Hi, how do you find the inverse of this function: $\displaystyle f(x) = \frac{15}{2x^2 + 11x +5}$ ?
Hello, sakonpure6!
How do you find the inverse of this function? .$\displaystyle f(x) = \frac{15}{2x^2 + 11x +5}$
Use the prescribed procedure.
[However, the result will not be a function.]
We have: .$\displaystyle y \;=\;\frac{15}{2x^2+11x+5}$
"Switch" variables: .$\displaystyle x \;=\;\frac{15}{2y^2 + 11y + 5}$
Solve for $\displaystyle y\!:$
. . $\displaystyle x(2y^2+11y + 5) \:=\:15 \quad\Rightarrow\quad 2xy^2 + 11xy + 5x - 15 \:=\:0$
Quadratic Formula: .$\displaystyle y \;=\;\dfrac{-11x \pm\sqrt{(11x)^2-4(2x)(5x-15)}}{2(2x)} $
. . . . . . . . . . . . . . . $\displaystyle y \;=\;\frac{-11x \pm \sqrt{121x^2 - 40x^2 + 120x}}{4x}$
. . . . . . . . . . . . . . . $\displaystyle y \;=\;\frac{-11x \pm\sqrt{81x^2 + 120x}}{4x}$
Therefore: .$\displaystyle f^{-1}(x) \;=\;\frac{-11x \pm \sqrt{81x^2+120x}}{4x}$
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).
$\displaystyle f^{-1}(x)=g(x)$ for x in the restricted domain you chose.