Question

what is the critical number?

f(x) = 19 xln x

Answer 1

my friend, that will take you quite a while

Answer 2

\[f'(x)=19\ln(x)+19\]

Answer 3

set
\[f'(x)=0\] get
\[\ln(x)+1=0\] so
\[\ln(x)=-1\implies x = e^{-1}=\frac{1}{e}\]

Answer 4

f(x) = 19 xln x
f'(x) = 19 ( lnx + x/x) = 19 ( lnx + 1)
f'(x ) = 0 --> lnx = -1 => x = e^-1 = 1/e