Find the maximum and the minimum values of the function $f\left(x\right) = 5-3x+5x^2 - x^3$, $x \in R$
$f\left(x\right) = 5-3x+5x^2-x^3$
$\therefore$ $f'\left(x\right) = -3 + 10x -3x^2$
For maxima or minima, $f'\left(x\right) = 0$
i.e. $-3+10x-3x^2 = 0$
i.e. $\left(3x-1\right)\left(x-3\right)=0$
$\therefore$ $f'\left(x\right) = 0$ $\implies$ $x=\dfrac{1}{3}$ or $x=3$
Now, $f''\left(x\right) = 10-9x$
$\therefore$ $f''\left(\dfrac{1}{3}\right) = 7 > 0$
$\implies$ f has local minimum at $x=\dfrac{1}{3}$ and maximum value $=f\left(\dfrac{1}{3}\right) = \dfrac{122}{27}$
$f''\left(3\right) = -17 < 0$
$\implies$ f has local maximum at $x=3$ and minimum value $= f\left(3\right) = 14$