Question

The range of the polynomial $p\left(x\right)=(4x^3-3x)$ as $x$ varies over the interval $(-\frac{1}{2},\frac{1}{2})$ is

• A. $[-1,1]$
• B. $(-1,1]$
• C. $(-1,1)$
• D. $(-\frac{1}{2},\frac{1}{2})$

Answer 1

Answer: (C)

$(-1,1)$

Given,

$p\left(x\right)=(4x^3-3x)$

$p^{\prime }\left(x\right)=12x^2-3$

$p^{\prime }\left(x\right)=12x^2-3=0=x^2=\frac{3}{12}=x=\sqrt{\frac{1}{4}}=\pm \frac{1}{2}$

$p^{\prime }\left(x\right)\le 0$ in the interval $(-\frac{1}{2},\frac{1}{2})$

Hence the $p\left(x\right)$ is a decreasing function over the given domain. Maximum value occurs at $x=-\frac{1}{2}$ and minimum at $x=\frac{1}{2}$

$p(-\frac{1}{2})=1$ & $p\left(\frac{1}{2}\right)=-1$

Range : $(-1,1)$