Question

Consider $f:R^{+}\to [-9,\infty ]$ given by $f\left(x\right)=5x^2+6x-9$. Prove that $f$ is invertible with $f^{\prime }\left(y\right)=\left(\frac{\sqrt{54+5y}-3}{5}\right)$.

Answer 1

The domain is restricted to positive real numbers only, so:

$y=f\left(x\right)=5x^2+6x-9=5({(x+\frac{3}{5})}^2-\frac{54}{25})\Rightarrow {(x+\frac{3}{5})}^2=\frac{54+5y}{25}\Rightarrow x=\frac{\sqrt{54+5y}-3}{5}$ for all $y$ in the range.

Thus the function is onto.

Now lets assume $x_1\ne x_2,$

then $y_1=y_2\Rightarrow 5({(x_1+\frac{3}{5})}^2-\frac{54}{25})=5({(x_2+\frac{3}{5})}^2-\frac{54}{25})\Rightarrow x_1=x_2$

which is a contradiction to our assumption and therefore $x_1=x_2.$

Thus the function is one-one.

Since the function is both one-one and onto, it is invertible on the given domain. Its inverse is given by the expression for $x$ found above while proving that the function is onto.