Find the points of local maxima, local minima and the points of infle…

The correct answer is Option (2) → Max at $x=1$ (val 0), Min at $x=3$ (val -28) ##

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

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

$f'(x) = 5x^4 - 20x^3 + 15x^2$

For maxima or minima, $f'(x) = 0$

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

$\Rightarrow 5x^2(x^2 - 4x + 3) = 0$

$\Rightarrow 5x^2(x^2 - 3x - x + 3) = 0$

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

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

$∴x = 0, 1, 3$

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

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

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

$∴$ Maximum value of $y = 1 - 5 + 5 - 1 = 0$

and minimum value $= (3)^5 - 5(3)^4 + 5(3)^3 - 1$

$= 243 - 81 \times 5 + 27 \times 5 - 1$

$= -28$