Find the stationary points of the function $f(x)=3x^4-8x^3+6x^2$ and distinguish between them. Also find the local maximum and local minimum values, if they exist.

Stationary points at $x=0$ (local minimum) and $x=1$ (inflection point). Local minimum value is 0. No local maximum.

The correct answer is Option (2) → Stationary points at $x=0$ (local minimum) and $x=1$ (inflection point). Local minimum value is 0. No local maximum.

Given $f(x) = 3x^4-8x^3+6x^2$.

It being a polynomial function is differentiable for all $x ∈ R$.

Diff. it w.r.t. x, we get

$f'(x) = 3.4x^3-8.3x^2+6.2x = 12 (x^3 - 2x^2 + x)$ and

$f''(x) = 12(3x^2-4x+1)$.

For stationary points $f'(x) = 0⇒ 12(x^3-2x^2 + x) = 0$

$⇒x(x^2-2x+1)=0⇒x(x-1)^2=0⇒x=0,1$.

At $x=0$

$f''(0) = 12.(3.0-4.0+ 1) = 12 > 0$

⇒ f has a local minima at x = 0 and local minimum value at x = 0 is

$f(0) = 3.0-8.0 +6.0 = 0$.

At $x = 1$

$f''(1)=12(3.1^2-4.1+1)= 12.0=0,$

which does not give any inference, so we calculate $f'''(1)$

$f'''(x)=\frac{d}{dx}(12(3x^2-4x+1)) = 12. (6x-4)$

$∴ f'''(1) = 12.(6.1-4) = 24 ≠0$

⇒ f has a point of inflexion and

$f(1) = 3.1^4-8.1^3 +6.1^2=3-8+6=1$.

Hence, the stationary points of the given function are $x = 0, 1; x = 0$ is a point of local minima and local minimum value = 0; $x = 1$ is a point of inflexion and inflexional value = 1. The local minimum value of the function is 0 and it has no maximum value.