Question

Locate the critical points of the following functions and use the Second Derivative Test to determine (if possible) whether they correspond to local maxima or local minima. $$p(t)=2 t^{3}+3 t^{2}-36 t$$

Short answer

Question: Determine the critical points of the function $$p(t) = 2t^3 + 3t^2 - 36t$$ and use the Second Derivative Test to identify whether they correspond to local maxima or local minima. Answer: The critical points are t = -3 and t = 2. The Second Derivative Test shows that t = -3 corresponds to a local maximum and t = 2 corresponds to a local minimum.

Step-by-step solution

Find the first derivative of the function

Differentiate the given function $$p(t)=2 t^{3}+3 t^{2}-36 t$$ with respect to t. The first derivative is: $$p'(t)=\frac{d}{dt}(2t^3+3t^2-36t)$$ Using the power rule, we get: $$p'(t)=6t^2+6t-36$$

Find the critical points

To find the critical points, we need to solve the equation $$p'(t)=0$$. So, we set the first derivative equal to zero and solve for t: $$6t^2+6t-36=0$$ Factor out the greatest common divisor (GCD) among the coefficients which is 6: $$6(t^2+t-6)=0$$ Divide both sides by 6: $$t^2+t-6=0$$ Factor the quadratic equation: $$(t+3)(t-2)=0$$ Now find the values of t: $$t=-3, 2$$ So the critical points are t=-3 and t=2.

Find the second derivative of the function

Differentiate the first derivative $$p'(t)=6t^2+6t-36$$ with respect to t. The second derivative is: $$p''(t)=\frac{d}{dt}(6t^2+6t-36)$$ Using the power rule, we get: $$p''(t)=12t+6$$

Determine the nature of critical points using the Second Derivative Test

We will now use the Second Derivative Test to determine if the critical points correspond to local maxima or local minima. Evaluate the second derivative at the critical points: For t = -3: $$p''(-3)=12(-3)+6=-30$$ Since p''(-3) is negative, the critical point t=-3 corresponds to a local maximum. For t = 2: $$p''(2)=12(2)+6=30$$ Since p''(2) is positive, the critical point t=2 corresponds to a local minimum. So, by the Second Derivative Test, the critical point t=-3 corresponds to a local maximum, and the critical point t=2 corresponds to a local minimum.

Key concepts

First Derivative

Finding the first derivative is crucial when analyzing critical points of a function. For the function \(p(t) = 2t^3 + 3t^2 - 36t\), we must differentiate with respect to \(t\). This process identifies how the function \(p(t)\) is changing at any point. Using the power rule (which involves multiplying the power by the coefficient and subtracting one from the power), we obtain:

• \(p'(t) = 6t^2 + 6t - 36\)

To find where the function changes direction, we set \(p'(t)\) equal to zero. Solving \(6t^2 + 6t - 36 = 0\) provides us with the critical points of the function. These points are when the slope or gradient of the function is zero, indicating a potential local maximum or minimum.

Second Derivative Test

The Second Derivative Test helps us determine whether each critical point is a local maximum, minimum, or neither. After finding the first derivative, \(p'(t)\), we differentiate it again to find the second derivative, \(p''(t)\). The formula here is:

• \(p''(t) = 12t + 6\)

This expression describes the concavity of the original function. Evaluating \(p''(t)\) at each critical point gives insight into the function's curvature at those points:

• If \(p''(t) > 0\), the function is concave up, and the critical point is a local minimum.
• If \(p''(t) < 0\), the function is concave down, and the critical point is a local maximum.
• If \(p''(t) = 0\), the test is inconclusive.

Local Maximum

A local maximum is a point where the function reaches a peak just within a nearby region. To identify these points using the Second Derivative Test, we look at critical points where \(p''(t) < 0\). For the function \(p(t) = 2t^3 + 3t^2 - 36t\), and the critical point \(t = -3\):

• \(p''(-3) = 12(-3) + 6 = -30\)

Since \(p''(-3) < 0\), the function is concave down at this point, confirming that \(t = -3\) is a local maximum. This means the value of \(p(t)\) at \(t = -3\) is higher than any nearby points, creating a peak in the function's graph.

Local Minimum

A local minimum occurs where the function takes on its lowest value relative to nearby points. We use the Second Derivative Test to identify these points, where \(p''(t) > 0\). For our function, at the critical point \(t = 2\):

• \(p''(2) = 12(2) + 6 = 30\)

Since \(p''(2) > 0\), the function is concave up at \(t=2\), confirming a local minimum. This implies the value of \(p(t)\) at \(t = 2\) is lower than nearby points, creating a trough in the graph. Thus, \(t = 2\) represents a local minimum for the function \(p(t)\).