Questions: We obtained that p'(t) = 6t^2 - 48t + 90. Find the critical points of the function p(t) by setting p'(t) = 0 and solving for t. Separate multiple answers with a comma.

We obtained that p'(t) = 6t^2 - 48t + 90. Find the critical points of the function p(t) by setting p'(t) = 0 and solving for t. Separate multiple answers with a comma.
Transcript text: We obtained that $\mathrm{p}^{\prime}(\mathrm{t})=6 \mathrm{t}^{2}-48 \mathrm{t}+90$. Find the critical points of the function $\mathrm{p}(\mathrm{t})$ by setting $\mathrm{p}^{\prime}(\mathrm{t})=0$ and solving for t . Separate multiple answers with a comma.
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Solution

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Solution Steps

Step 1: Set the derivative equal to zero.

We are given $p'(t) = 6t^2 - 48t + 90$. To find the critical points, we set the derivative equal to zero: $6t^2 - 48t + 90 = 0$

Step 2: Solve for t.

Divide the equation by 6: $t^2 - 8t + 15 = 0$ Factor the quadratic equation: $(t - 3)(t - 5) = 0$ This gives us two solutions for t: $t = 3$ or $t = 5$

Final Answer

The critical points are t = 3, 5.

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