Question

If g(x) = x² + x + 1 and (g o f)(x) = 4x² - 10x + 5. Then find f(5/4).​

Answer 1

$\Large\boxed{\text{The value does not exist.}}$

$\Large\text{\underline{\underline{Exact form}}}$

Let us find the exact form of the equation.

Let us suppose that -

$\cdots\longrightarrow f\left(x\right)=t.$

Then we get the following equations -

$\cdots\longrightarrow\begin{cases} & (g\circ f)(x)=4x^{2}-10x+5 \\ & g(t)=t^{2}+t+1. \end{cases}$

Equating,

$\cdots\longrightarrow 4x^{2}-10x+5=t^{2}+t+1$

$\cdots\longrightarrow t^{2}+t-4x^{2}+10x-4=0$

$\text{ \cdots\longrightarrow t=\dfrac{-1\pm\sqrt{1+16x^{2}-40x+16}}{2} }$

$\text{ \cdots\longrightarrow t=\dfrac{-1\pm\sqrt{16x^{2}-40x+17}}{2} }$

Hence, -

$\text{ \cdots\longrightarrow\boxed{f(x)=\dfrac{-1+\sqrt{16x^{2}-40x+17}}{2}} }$

or, -

$\text{ \cdots\longrightarrow\boxed{f(x)=\dfrac{-1-\sqrt{16x^{2}-40x+17}}{2}.} }$

The domain is restricted to -

$\text{ \cdots\longrightarrow \boxed{x\in\mathbb{R}-(\dfrac{5-\sqrt{2}}{4},\dfrac{5+\sqrt{2}}{4}).} }$

$\Large\text{\underline{\underline{Explanation}}}$

We know that the domain of the function is restricted to -

$\text{ \cdots\longrightarrow \boxed{x\in\mathbb{R}-(\dfrac{5-\sqrt{2}}{4},\dfrac{5+\sqrt{2}}{4}).} }$

But, -

$\text{ \cdots\longrightarrow x=\dfrac{5}{4} is not in the domain.}$

Hence, -

$\text{ \cdots\longrightarrow the value of f\left(\dfrac{5}{4}\right) \underline{does not exist.}}$