Application of Derivatives — Class 12 Maths Solution

Given that, $f(x) = {x^5} - 5{x^4} + 5{x^3} - 1$

On differentiating w.r.t. $x$, we get
${f^\prime }(x) = 5{x^4} - 20{x^3} + 15{x^2}$

For maxima or minima, ${f^\prime }(x) = 0$

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

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

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

$\Rightarrow$ $5{x^2}[x(x - 3) - 1(x - 3)] = 0$

$\Rightarrow$ $5{x^2}[(x - 1)(x - 3)] = 0$
$\therefore x = 0,1,3$

Sign scheme for $\frac{{dy}}{{dx}} = 5{x^2}(x - 1)(x - 3)$

So, $y$ has maximum value at $x = 1$ and minimum value at $x = 3$.

At $x = 0,y$ has neither maximum nor minimum value.

Therefore the maximum value of $y = 1 - 5 + 5 - 1 = 0$

and minimum value $= {(3)^5} - 5{(3)^4} + 5{(3)^3} - 1$
$= 243 - 81 \times 5 - 27 \times 5 - 1 = - 298$