For the function $f(x) = -2x^3 + 3x^2 + 36x - 10$, which of the following is/are true?

(A) $f$ is increasing in $(-\infty, -2)$

(B) $f$ is increasing in $(-2, 3)$

(C) $f$ is decreasing in $(-\infty, -2)$

(D) $f$ is decreasing in $(3, \infty)$

Choose the correct answer from the options given below:

We need to find where the function $f(x) = -2x^3 + 3x^2 + 36x - 10$ is increasing (going upward) or decreasing (going downward).

The Secret: A function is increasing when its derivative is positive and decreasing when its derivative is negative.

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Find the derivative using the power rule (bring down the exponent and reduce the power by 1):

$f'(x) = -6x^2 + 6x + 36$

Why? $-2x^3$ becomes $-6x^2$, $3x^2$ becomes $6x$, $36x$ becomes $36$, and the constant $-10$ disappears.

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Simplify by factoring out $-6$:

$f'(x) = -6(x^2 - x - 6)$

Now factor the quadratic. We need two numbers that multiply to $-6$ and add to $-1$. Those are $-3$ and $+2$.

$f'(x) = -6(x - 3)(x + 2)$

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Find critical points where $f'(x) = 0$:

$-6(x - 3)(x + 2) = 0$

This gives us: $x = 3$ and $x = -2$

These points divide our number line into three regions: $(-\infty, -2)$, $(-2, 3)$, and $(3, \infty)$

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For interval $(-\infty, -2)$, test $x = -3$:

$f'(-3) = -6(-3 - 3)(-3 + 2) = -6(-6)(-1) = -36$

Negative! So $f$ is decreasing on $(-\infty, -2)$

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For interval $(-2, 3)$, test $x = 0$:

$f'(0) = -6(0 - 3)(0 + 2) = -6(-3)(2) = 36$

Positive! So $f$ is increasing on $(-2, 3)$

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For interval $(3, \infty)$, test $x = 4$:

$f'(4) = -6(4 - 3)(4 + 2) = -6(1)(6) = -36$

Negative! So $f$ is decreasing on $(3, \infty)$

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(A) $f$ is increasing in $(-\infty, -2)$ - FALSE (it's decreasing)

(B) $f$ is increasing in $(-2, 3)$ - TRUE

(C) $f$ is decreasing in $(-\infty, -2)$ - TRUE

(D) $f$ is decreasing in $(3, \infty)$ - TRUE

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Therefore, statements (B), (C), and (D) are correct.