Find the intervals in which the function
$f(x) = 2x^3 - 3x^2 - 36x + 7$
is:
(a) Strictly increasing
(b) Strictly decreasing

Q: Find the intervals in which the function $f(x) = 2x^3 - 3x^2 - 36x + 7$ is: (a) Strictly increasing (b) Strictly decreasing

Explanation: Differentiate the function: \[f'(x) = \dfrac{d}{dx}(2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36\] Factor the derivative: \[f'(x) = 6(x^2 - x - 6) = 6(x + 2)(x - 3)\] Set $f'(x) = 0$ to find critical points: \[6(x + 2)(x - 3) = 0 \Rightarrow x = -2,,3\] Now analyze the sign of $f'(x)$ in the intervals: \[\] In $(-\infty, -2)$: Both $(x + 2) 0$ $\Rightarrow f(x)$ is strictly increasing on $(-\infty, -2)$ \[\] In $(-2, 3)$: $(x + 2) > 0$ and $(x - 3) 0$ and $(x - 3) > 0 \Rightarrow f'(x) > 0$ $\Rightarrow f(x)$ is strictly increasing on $(3, \infty)$ \[\] Final Answer: $f(x)$ is strictly increasing on $(-\infty, -2)$ and $(3, \infty)$ and\[\] $f(x)$ is strictly decreasing on $(-2, 3)$

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Explanation: Differentiate the function: $$f'(x) = \dfrac{d}{dx}(2x^3 - 3x^2 - 36x + 7) = 6x^2 - 6x - 36$$ Factor the derivative: $$f'(x) = 6(x^2 - x - 6) = 6(x + 2)(x - 3)$$ Set $f'(x) = 0$ to find critical points: $$6(x + 2)(x - 3) = 0 \Rightarrow x = -2,,3$$ Now analyze the sign of $f'(x)$ in the intervals: In $(-\infty, -2)$: Both $(x + 2) < 0$ and $(x - 3) < 0 \Rightarrow f'(x) > 0$ $\Rightarrow f(x)$ is strictly increasing on $(-\infty, -2)$ In $(-2, 3)$: $(x + 2) > 0$ and $(x - 3) < 0 \Rightarrow f'(x) < 0$ $\Rightarrow f(x)$ is strictly decreasing on $(-2, 3)$ In $(3, \infty)$: Both $(x + 2) > 0$ and $(x - 3) > 0 \Rightarrow f'(x) > 0$ $\Rightarrow f(x)$ is strictly increasing on $(3, \infty)$ Final Answer: $f(x)$ is strictly increasing on $(-\infty, -2)$ and $(3, \infty)$ and $f(x)$ is strictly decreasing on $(-2, 3)$

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