Find the points of local maxima,

Given, f (x) = x5 – 5x4 + 5x3 – 1

Differentiating the function,

f’ (x) = 5x4 – 20x3 + 15x2

For local maxima and local minima, f’ (x) = 0

$5 x^{4}-20 x^{3}+15 x^{2}=0 \Rightarrow 5 x^{2}\left(x^{2}-4 x+3\right)=0$

$\Rightarrow 5 x^{-2}\left(x^{2}-3 x-x+3\right)=0 \Rightarrow x^{2}(x-3)(x-1)=0$

$\therefore x=0, x=1$ and $x=3$

$\Rightarrow \quad f^{\prime \prime}(x)_{\text {at } x=0}=20(0)^{3}-60(0)^{2}+30(0)=0$ which is neither

maxima nor minima.

$\therefore f(x)$ has the point of inflection at $x=0$

$f^{\prime \prime}(x)_{\text {at } x-1}=20(1)^{3}-60(1)^{2}+30(1)$

$=20-60+30=-10<0$ Maxima

$f^{\prime \prime}(x)_{\mathrm{at} x-3}=20(3)^{3}-60(3)^{2}+30(3)$

$=540-540+90=90>0$ Minima

The maximum value of the function at x = 1

f(x) = (1)5 – 5(1)4 + 5(1)3 – 1

= 1 – 5 + 5 – 1 = 0

The minimum value at x = 3 is

f(x) = (3)5 – 5(3)4 + 5(3)3 – 1

= 243 – 405 + 135 – 1

= 378 – 406 = -28

Therefore, the function has its maxima at x = 1 and the maximum value = 0 and its has minimum value at x = 3 and its minimum value is -28.