Let f(x)=e^(3x)-kx, for k>0.

Question

Using a calculator or computer sketch the graph of f for k=1/9,1/6,1/3,1/2,1,2,4. Describe what happens as k changes.
1.f(x) has a local minimum. Find the location of the minimum.
2. Find the y-coordinate of the minimum.
3. Find the value of k for which this y-coordinate is the largest.
4. How do you know that this value of k maximizes the y-coordinate? Find d^2y/dk^2 to use the second-derivative test.

Answer 1

f' = 3e^(3x) - k
so it has a minimum when x = 1/3 ln(k/3)

to find k where the y-coordinate is largest, note that this coordinate is
e^(3(1/3 ln(k/3)) - k(1/3 ln(k/3))
= e^(ln(k/3)) - k/3 ln(k/3)
= k/3 (1 - ln(k/3))
which is a maximum at k=3

yo can check the 2nd derivative

Answer 2

For the second derivative, I got 9e^(3x) but it says its wrong. Am I missing a step?

Answer 3

Find a formula for a curve of the form y=e−(x−a)2/b
for b>0 with a local maximum at x=6 and points of inflection at x=3 and x=9.