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

Hint: 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.