y = e^(2x)-kx
y' = 2e^(2x) - k
y" = 4e^(2x)
so, now we can answer the questions
y'=0 at x = 1/2 ln(k/2)
y" is always positive, so
there is a minimum at (1/2 ln(k/2) , k/2 (1-ln(k/2))
Now, which value of k makes y=k/2 (1-ln(k/2) a maximum is k=2
since
y' = -1/2 ln(k/2)
y" = -1/2x < 0 at x=2
Let f(x)=e^(2x)-kx, for k greater than 0.
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.
f(x) has a local minimum. Find the location of the minimum.
x= ____
Find the y-coordinate of the minimum.
y= _____
Find the value of k for which this y-coordinate is largest.
k= ______
How do you know that this value of k maximizes the y-coordinate? Find d^2y/dk^2 to use the second-derivative test.
d^2y/d^2k=
(Note that the derivative you get is negative for all positive values of k, and confirm that you agree that this means that your value of k maximizes the y-coordinate of the minimum.)
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