Note: You can earn partial credit on this problem. Let , for . Using a calculato

Question

Note: You can earn partial credit on this problem.

Using a calculator or computer, sketch the graph of for . Describe what happens as changes.

has a local minimum. Find the location of the minimum.

Find the -coordinate of the minimum.

Find the value of for which this -coordinate is largest.

How do you know that this value of maximizes the -coordinate? Find to use the second-derivative test.

(Note that the derivative you get is negative for all positive values of , and confirm that you agree that this means that your value of maximizes the -coordinate of the minimum.)

Explanation / Answer

The function behaves like a line for negative x, and the slope of this line gets steeper as K increases. local min where f '(x) = 5e^(5x) - k = 0 so e^(5x) = k/5 so 5x = ln(k/5) so x = (1/5)ln(k/5) y = f((1/5)ln(k/5)) = k/5 - (k/5)ln(k/5) now take derivative of this last expression with respect to k to get 1/5 -(1/5)ln(k/5) -(k/5)(5/k)(1/5) = 1/5 - (1/5)ln(k/5) -1/5 = (1/5)ln(k/5); this equal 0 when ln(k/5) = 0, so k/5 = 1 and k = 5. The second derivative of y with respect to k is the derivative of (1/5)ln(k/5), which is (1/5)(5/k)(1/5) = k/5 When k = 5, k/5 = 1 >0, so y is concave up and this k = 5 corresponds to a min.

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.