Find all the points of local maxima and local minima of the function…

The correct answer is Option (2) → Local minima at $x=-3$, Local maxima at $x=0$ and $x=−5$

Given $f(x) = -\frac{3}{4}x^4-8x^3-\frac{45}{2}x^2 + 105$.

Clearly, $D_f = R$ and f is differentiable for all $x ∈ R$.

Differentiating w.r.t. x, we get

$f'(x)=-3x^3-24x^2 - 45x$ and

$f''(x)=-9x^2-48x-45$.

For critical points, $f'(x) = 0⇒-3x (x^2 + 8x + 15) = 0$

$⇒x(x+3)(x+5)=0⇒x= 0, -3, -5$.

Therefore, the points where the extremum may occur are $x = 0, -3, -5$.

At $x=0$

$⇒ f''(0) = -9.0-48.0-45=-45 < 0$

⇒ O is a point of local maxima.

At $x=-3$

$f''(-3)=-9.(-3)^2 - 48. (-3) - 45=-81 +144-45 = 18 > 0$

⇒ -3 is a point of local minima.

At $x=-5$

$f''(-5)=-9.(-5)^2 - 48. (-5) - 45=-225 + 240-45 = −30 < 0$

⇒ $x = -5$ is a point of local maxima.

Hence, points of local maxima are $x = 0, x = -5$ and point of local minima is $x = -3$.