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

Rate of change of $f\left(x\right)\text{ is }{f}^{\text{'}}\left(x\right)$

$=15x^4-15x^2+5=15\left(x^4-x^2+\frac{1}{3}\right)$

$=15\left({\left(x^2-\frac{1}{2}\right)}^2+\frac{1}{12}\right)\ge \frac{15}{12}=\frac{5}{4}$

The minimum vale of $f$ is attained at $x=+\frac{1}{\sqrt{2}}$ and equals $\frac{5}{4}$.