Question

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

Answer 1

To prove that $f$ is invertible with $f'(y) = (\dfrac{{\sqrt {54 + 5y} - 3}}{5})$, we will first need to find the inverse of the function. Since the function is as quadratic polynomial, we need to find the roots of the quadratic equation $f(x) = 5{x^2} + 6x - 9$ in the range $f:{R^ + } \to [ - 9,\infty ]$. The value of the root in this range will be the inverse of the function $f(x) = 5{x^2} + 6x - 9$.

Complete step by step answer:
A function $f(x) = 5{x^2} + 6x - 9$which is define under $f:{R^ + } \to [ - 9,\infty ]$
Let $f(x) = y$
$f(x) = 5{x^2} + 6x - 9 = y$
$\Rightarrow 5{x^2} + 6x - (9 + y) = 0$
This has now become a quadratic equation.
Compare the equation with $a{x^2} + bx + c = 0$.
We get $a = 5$, $b = 6$ and $c = - (9 + y)$.
Find the root of the equations from the following equation
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
$\Rightarrow x = \dfrac{{ - 6 \pm \sqrt {36 - 4 \times 5 \times (9 + y)} }}{{2 \times 5}}$
$\Rightarrow x = \dfrac{{ - 6\sqrt {{6^2} + 20(a + y)} }}{{10}}$
$\Rightarrow x = \dfrac{{ - 6 \pm \sqrt {226 + 20y} }}{{10}}$
Choose the correct root
Since, $f:{R^ \pm } > [ - 9,\infty ]$
This means that the value of the function, $f(x) = 5{x^2} + 6x - 9$can be a positive or negative real number in the range $[ - 9,\infty ]$ with the endpoint included.
But, $x$ is positive for$y \in [ - 9,\infty ]$
$\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {4(54 + 5y)} }}{{10}}$
$\Rightarrow x = \dfrac{{ - 3 \pm 2\sqrt {(54 + 5y)} }}{{10}}$
$\Rightarrow x = \dfrac{{ - 3 \pm \sqrt {(54 + 5y)} }}{5}$
$x = f'(y) = \dfrac{{\sqrt {54 + 5y} - 3}}{{10}}$
Since we found the value of x, we can say that, $f$Is invertible when, $f:{R^ \pm } > [ - 9,\infty ]$ and $f'(y) = \dfrac{{\sqrt {54 + 5y} - 3}}{{10}}$

Note:
This question can also be solved in the following manner. Given the function $f(x)$ we want to find the inverse function, $f'(x)$.
First, we will replace $f(x)$ with $y$. This is done to make the rest of the calculation process easier. Now, we will replace every $x$ with a $y$and replace every$y$with an $x$. Solve the equation from Step 2 for $y$. This is the step where mistakes are most often made so be careful with this step.
Replace $y$ with $f'(x)$. In other words, we’ve managed to find the inverse at this point. Please verify your work by checking that
$(fof')(x) = x$ and $(f'of)(x) = x$ are both true. This work can sometimes be messy making it easy to make mistakes so again be careful here.