For the function, $f(x) = \frac{-3}{4}x^4 - 8x^3 - \frac{45}{2}x^2 - 350$, which of the following statements are correct?

(A) $x = -3$ and $x = -5$ are the only critical points of the given function.

(B) $x = -3$ is a point of local minimum.

(C) The local minimum value at $x = -3$ is 23.1.

(D) $x = -5$ is a point of local maximum.

Choose the correct answer from the options given below:

Given:

$f(x) = \dfrac{-3}{4}x^4 - 8x^3 - \dfrac{45}{2}x^2 - 350$

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Differentiating:

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

Setting $f'(x) = 0$:

$-3x^3 - 24x^2 - 45x = 0$

$-3x(x^2 + 8x + 15) = 0$

$-3x(x + 3)(x + 5) = 0$

$x = 0, \quad x = -3, \quad x = -5$

So the critical points are $x = 0$, $x = -3$, and $x = -5$.

Statement (A) claims $x = -3$ and $x = -5$ are the only critical points — this is incorrect since $x = 0$ is also a critical point.

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To classify the critical points, we find the second derivative:

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

At $x = -3$:

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

Since $f''(-3) > 0$, $x = -3$ is a point of local minimum.

Statement (B) is correct.

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At $x = -5$:

$f''(-5) = -9(25) - 48(-5) - 45 = -225 + 240 - 45 = -30 < 0$

Since $f''(-5) < 0$, $x = -5$ is a point of local maximum.

Statement (D) is correct.

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Finding the value of $f(-3)$:

$f(-3) = \dfrac{-3}{4}(81) - 8(-27) - \dfrac{45}{2}(9) - 350$

$= \dfrac{-243}{4} + 216 - \dfrac{405}{2} - 350$

$= -60.75 + 216 - 202.5 - 350$

$= -397.25$

Statement (C) claims the local minimum value at $x = -3$ is $23.1$ — this is incorrect since $f(-3) = -397.25$.

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Only statements (B) and (D) are correct.