If $f : R \rightarrow R$ is defined by $f(x) = x^{2} - 3x + 2$,find $f(f(x))$.

Vedclass pdf generator app on play store
Vedclass iOS app on app store
Given $f(x) = x^{2} - 3x + 2$.
To find $f(f(x))$,we substitute $f(x)$ into the function $f$:
$f(f(x)) = f(x^{2} - 3x + 2)$
$= (x^{2} - 3x + 2)^{2} - 3(x^{2} - 3x + 2) + 2$
$= (x^{4} + 9x^{2} + 4 - 6x^{3} + 4x^{2} - 12x) - 3x^{2} + 9x - 6 + 2$
$= x^{4} - 6x^{3} + (9x^{2} + 4x^{2} - 3x^{2}) + (-12x + 9x) + (4 - 6 + 2)$
$= x^{4} - 6x^{3} + 10x^{2} - 3x$

Explore More

Similar Questions

Let $f(x) = \frac{ax + b}{cx + d}$. Then,$f \circ f(x) = x$ provided that

Let the functions $f: R \rightarrow R$ and $g: R \rightarrow R$ be defined as $f(x) = \begin{cases} x+2, & x < 0 \\ x^2, & x \geq 0 \end{cases}$ and $g(x) = \begin{cases} x^3, & x < 1 \\ 3x-2, & x \geq 1 \end{cases}$. Then,the number of points in $R$ where $(f \circ g)(x)$ is $NOT$ differentiable is equal to

If $f$ is an increasing function and $g$ is a decreasing function, and $fog$ is defined, then what kind of function is $fog$?

If $f(x) = 3x + 10$ and $g(x) = x^2 - 1$,then $(fog)^{-1}(x) = $

If $f(x) = \begin{cases} 2+2x, & -1 \leq x < 0 \\ 1-\frac{x}{3}, & 0 \leq x \leq 3 \end{cases}$ and $g(x) = \begin{cases} -x, & -3 \leq x \leq 0 \\ x, & 0 < x \leq 1 \end{cases}$,then the range of $(f \circ g)(x)$ is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo