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(A) Step $1$: Find the first derivative of the function $f(t) = -2t^3 + 3t^2 + 5$ with respect to $t$.$f'(t) = \frac{d}{dt}(-2t^3 + 3t^2 + 5) = -6t^2

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$1 	ext{ hour}$

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(A) Step $1$: Find the first derivative of the function $f(t) = -2t^3 + 3t^2 + 5$ with respect to $t$.$f'(t) = \frac{d}{dt}(-2t^3 + 3t^2 + 5) = -6t^2 + 6t$.Step $2$: Set the first derivative to zero to find the critical points.$-6t^2 + 6t = 0 \Rightarrow -6t(t - 1) = 0$.This gives $t = 0$ and $t = 1$.Step $3$: Find the second derivative to determine the nature of the critical points.$f''(t) = \frac{d}{dt}(-6t^2 + 6t) = -12t + 6$.Step $4$: Evaluate the second derivative at the critical points.For $t = 1$, $f''(1) = -12(1) + 6 = -6$.Since $f''(1) < 0$, the function has a local maximum at $t = 1$ hour.

$A$ YouTube short video is going viral according to the function $f(t) = -2t^3 + 3t^2 + 5$. At what time $t$ (in hours) does the video get the maximum number of shares?

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