Question

The rate of change of the function $f\left(x\right)=3x^5-5x^3+5x-7$ is minimum when

• A. $x=0$
• B. $x=\frac{1}{\sqrt{2}}$
• C. $x=-\frac{1}{\sqrt{2}}$
• D. $x=\pm \frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $x=\pm \frac{1}{\sqrt{2}}$

Let
$f^{\prime }\left(x\right)=h\left(x\right)$
$=15x^4-15x^2+5$
Now
$h^{\prime }\left(x\right)=60x^3-30x=0$
$30x(2x^2-1)=0$
Or
$x=0$ and $x=\pm \frac{1}{\sqrt{2}}$.
Hence
$h(x=0)=5$ ..(i)
And
$h(x=\frac{1}{\sqrt{2}})$
$=h(x=-\frac{1}{\sqrt{2}})$
$=\frac{5}{4}$
Hence $f^{\prime }\left(x\right)$ is minimum at $x=\frac{\pm 1}{\sqrt{2}}$.