Question

If $f\left(x\right)=\{\begin{array}{cc}{x}^3,& x<0\\ 3x-2,& 0\le x\le 2\\ x^2+1,& x>2\end{array}$
Then find the value(s) of x for which f(x)=2.

• A. $2^{\frac{1}{3}}$
• B. $\frac{4}{3}$
• C. 1
• D. all of these

Answer 1

The correct option is B $\frac{4}{3}$

$f\left(x\right)=2f\left(x\right)=x^3,x<0x^3=2\Rightarrow x=2^{\frac{1}{3}}>0-notpossible.$
$f\left(x\right)=3x-2,0\le x\le 2f\left(x\right)=2\Rightarrow x=\frac{4}{3}0\le \frac{4}{3}\le 2-possible.$
$f\left(x\right)=x^2+1,x>2x^2+1=2\Rightarrow x^2=1\Rightarrow x=\pm 1.1<2\&-1<2-notpossible.$
Hence f(x) = 2 only when $x=\frac{4}{3}$