Can't do it until you show what e5x means
is it:
e^(5x) or (e^5)(x) or e(5x) or ...
Let f(x)=e5x−kx, for k>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.
2 answers
assuming the usual carelessness with parentheses, I'll go with
y = e^(5x)-kx
y' = 5e^(5x) - k
y" = 25e^(5x)
so, now we can answer the questions
y'=0 at x = 1/5 ln(k/5)
y" is always positive, so
there is a minimum at (1/5 ln(k/5) , k/5 (1-ln(k/5))
Now, which value of k makes y=k/5 (1-ln(k/5)) a maximum is k=5
since
y' = -1/5 ln(k/5)
y" = -1/5k < 0 at k=5
y = e^(5x)-kx
y' = 5e^(5x) - k
y" = 25e^(5x)
so, now we can answer the questions
y'=0 at x = 1/5 ln(k/5)
y" is always positive, so
there is a minimum at (1/5 ln(k/5) , k/5 (1-ln(k/5))
Now, which value of k makes y=k/5 (1-ln(k/5)) a maximum is k=5
since
y' = -1/5 ln(k/5)
y" = -1/5k < 0 at k=5
0 answers