y = 3x 3^x - 1
y' = 3*3^x + 3x(ln3)3^x
= 3*3^x * (1+x ln3)
So, y'=0 when x = -1/ln3
That is the only extremum. Now find whether y" is positive or negative there to determine which kind.
Can someone please explain how to solve this type of question?
Determine the maximum and minimum value of the function f(x)=3x3^x - 1
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