Question

what is the critical number of 8x ln x

Answer 1

what does critical number means?

Answer 2

it is a value of x for f'(x)=0 @suroj..as far as i know

Answer 3

take the derivative (using the product rule) set it equal to zero and solve for x

Answer 4

d/dx lnx= 1/x
knowing this, you can use the product rule
8x(1/x)+8lnx
8(1+lnx)=0
1+lnx=0
lnx=-1
e^-1=x

Answer 5

\[f'(x)=\frac{8x}{x}+8\ln(x)=8+8\ln(x)\]
\[8(1+\ln(x))=0\]
\[\ln(x)=-1\]
\[x=e^{-1}\]

Answer 6

what dockworker said

Answer 7

is this even question of derivative?

Answer 8

what about the function x^(4/5) (x- 10)

it has two critical numbers where A<B

what are A and B

Answer 9

i think you asked this earlier. didn't someone solve it?

Answer 10

critical number also occurs when the derivative is not defined if this helps you

Answer 11

ln(x) is not defined when x=0

Answer 12

i dont remember asking this earlier

Answer 13

maybe i did. i have a boatload of questions

Answer 14

\[\frac{d}{dx}x^{\frac{4}{5}}(x-10)=\frac{4}{5}x^{\frac{-1}{5}}(x-10)+x^\frac{4}{5}\]
\[=\frac{4}{5x^{\frac{1}{5}}}(x-10)+x^{\frac{4}{5}}\]
\[=\frac{4(x-10)+5x^{\frac{1}{5}}(x^{\frac{4}{5}})}{5x^{\frac{1}{5}}}\]
\[=\frac{4x-40+5x}{5x^{\frac{1}{5}}}=\frac{9x-40}{5x^{\frac{1}{5}}}\]
so the 2 critical numbers are x=40/9 and x=0, assuming i made no mistakes