Question

find the critical points(solve for x) for the derivative of the function.

f(x)= 5ln((x^2)+1)-3x

Answer 1

\[f'(x)=5 \cdot \frac{(x^2+1)'}{(x^2+1)}-3\]
you mean find the critical numbers of using f'

Answer 2

you mean find the critical numbers of f using f'*

Answer 3

\[5 \cdot \frac{2x}{x^2+1}-3=0\]
solve for x

Answer 4

\[\frac{10}{x^2+1}-\frac{3(x^2+1)}{x^2+1}=\frac{10-3x^2-3}{x^2+1}=\frac{-3x^2+7}{x^2+1}=0\]

also f' exist every where

so the only critical numbers are when
\[-3x^2+7=0\]

Answer 5

\[3x^2=7=> x^2=\frac{7}{3}=> x=\pm \sqrt{\frac{7}{3}}\]

Answer 6

thanks you so much :)

Answer 7

ninja i made a mistake

Answer 8

\[\frac{10x}{x^2+1}-\frac{3(x^2+1)}{x^2+1}=\frac{10x-3x^2-3}{x^2+1}=\frac{-3x^2+10x-3}{x^2+1}=0\]

Answer 9

\[-3x^2+10x-3=0 =>3x^2-10x+3=0=>3x^2-9x-x+3=0\]
\[3x(x-3)-1(x-3)=(x-3)(3x-1)=0=> x=3 or x=\frac{1}{3}\]

Answer 10

its cool, yeah that answer reflects the answers on the back of the book. thank you

Answer 11

i left off my x when i went to the next step

Answer 12

did you see that?

Answer 13

yeah