Let f(x) = - 2x^ 3 + 21x^ 2 - 60x + 41 , then

Given, $f\left(x\right) = - 2x^{3} + 21x^{2} - 60x + 41$

On differentiating w.r.t. $x$, we get

$f^{'} \left(x\right) = - 6x^{2} + 42x - 60$

$⇒ f^{'}$\left(x\right) = - 6$\left(x^{2} - 7x + 10\right)$

$⇒ f^{'} \left(x\right) = - 6 \left(x - 2\right)\left(x - 5\right)$

$⇒ x 5, f^{'} \left(x\right) < 0$ i.e., $f\left(x\right)$ is decreasing

and $2 0$ i.e., $f \left(x\right)$& is increasing