Find the turning points on the curve y = x^3-2x^2-4x-1 and distinguis…

The correct answer is Option (2) → Turning points at $x = -\frac{2}{3}$ (max), $x=2$ (min); local max = $\frac{13}{27}$, local min = −9

Let $y = f(x)$, then $f(x) = x^3-2x^2-4x-1$.

It being a polynomial function is differentiable for all $x ∈ R$, diff. it w.r.t. x, we get

$f'(x) = 3x^2-2.2x-4.1+0=3x^2-4x-4$ and

$f''(x) = 3.2x-4.1=6x-4$.

For turning points, $f'(x) = 0⇒ 3x^2-4x-4=0$

$⇒(x-2) (3x+2)=0⇒x=2,-\frac{2}{3}$

Therefore, the points where the extremum may occur are $x=2,-\frac{2}{3}$.

At $x = 2$

$f''(x) = 6.2-4=8>0$

⇒ f has a local minima at $x = 2$ and local minimum value at $x = 2$ is

$f(2) = 2^3-2.2^2-4.2-1-8-8-8-1=-9$.

At $x=-\frac{2}{3}$

$f''(-\frac{2}{3})=6.(-\frac{2}{3})-4=-4-4=-8<0$

$=-\frac{8}{27}-\frac{8}{9}+\frac{8}{3}-1=\frac{13}{27}$

Hence, the turning points of the given curve are (2, −9) and $(-\frac{2}{3},\frac{13}{27})$; (2, -9) is a point of minima and $(-\frac{2}{3},\frac{13}{27})$ is a point of maxima. The local maximum value of the function is $\frac{13}{27}$ and its local minimum value is -9.