Find the points of local maxima and minima of the function $f(x) = 3x^4 -4x^3 + 5$ in [-1, 2]. Also find absolute maximum and minimum values.

Absolute maximum value = 21 at $x=2$; Absolute minimum value = 4 at $x=1$.

The correct answer is Option (1) → Absolute maximum value = 21 at $x=2$; Absolute minimum value = 4 at $x=1$.

Given $f(x) = 3x^4 - 4x^3 +5, x ∈ [-1, 2]$   ...(i)

Clearly f(x) is differentiable for all $x ∈ [-1, 2]$.

Differentiating (i) w.r.t. x, we get

$f'(x)=12x^3-12x^2$ and

$f''(x) = 36x^2 - 24x$.

Now $f'(x) = 0⇒12x^3 - 12x^2 = 0⇒ x^2 (x − 1) = 0$

$⇒ x = 0, 1$ and both $0, 1 ∈ [-1, 2]$.

Therefore, the points where extremum may occur are

(i) $x = 0, x = 1$ where $f'(x) = 0$

(ii) $x = -1$, the left end point of [−1, 2]

(iii) $x = 2$, the right end point of [-1, 2].

At $x=0$

$f''(0) = 36.0-24.0=0,$

which does not give any inference, so we calculate $f'''(0)$.

$f'''(x) =\frac{d}{dx} (36x^2-24x) = 72x-24⇒f'''(0) = -24≠0$

⇒ f has neither maxima nor minima at $x = 0$ i.e. $x = 0$ is a point of inflexion.

At $x=1$

$f''(1) = 36.1^2-24.1 = 12 > 0$

⇒ f has a local minima at $x = 1$.

Local minimum value = $f(1) = 3.1^4 - 4.1^3 +5 = 4$.

Left end point $x=-1$

When $x > -1$ slightly, then

$f'(x)= 12 (-ve) -12 (+ve) = -ve$

⇒ f has a local maxima at $x = -1$.

Local maximum value = $f(-1) = 3.(-1)^4 - 4.(-1)^3 + 5 = 12$.

Right end point $x = 2$

When $x < 2$ slightly, then

$f'(x) = 12 (\text{slightly <8)-12 (slightly < 4) = +ve}$

⇒ f has a local maxima at $x = 2$.

Local maximum value = $f(2) = 3.2^4-4.2^3+5=21$.

Thus, f has a local minima at $x = 1$, and local maxima at $x = -1,2$.

Absolute maximum value = 21 and absolute minimum value = 4.