Can anyone provide any evidence for this calculus question? Please help.

Question

Explain why the function y = e^x + e^2x has no local extrema.

Answer 1

Explain why the function y = e^x + e^2x has no local extrema.

Detailed Solution:

To find local extrema, try to solve: dy/dx = 0. If there is no solution, means: no local extrema

dy/dx = e^x + 2e^2x

dy/dx = 0 ==> e^x + 2e^2x = 0 ==> 2e^2x = – e^x ==> 2e^x = – 1 ==> e^x = – 1/2 ==> ln e^x = ln(– 1/2)



==> x ln e = ln(– 1/2) ==> x = ln(– 1/2) = DNE (Does Not Exist = No Solution)

Therefore, No Solution, means: No Local Extrema

Answer 2

to find local extema take the derivative, set = 0, and solve for x

y' = e^x +2e^x = 0

3e^x = 0

e^x = 0

but no value of x will ever make e^x = 0

e^x = 0

take natural logs

x = ln(0) but ln0 is undefined, It's negative infinity which is not a real number.

y is everywhere increasing. y' is never zero. y' is always positive.

the local extrema would be where y'=0 which never happens