Question

Calculus
Find critical numbers of the function
f(x) 4x/ x^2+1

Answer 1

\[f(x)\frac{ 4x }{ x^2+1 }\]
\[f'(x)=\frac{4(x^2+1)-[4x(2x)] }{ (x^2+1)^2 }\]

Answer 2

Looks good so far.
You can cancel out from the numerator and denominator i think.

Answer 3

(x^2 + 1)^2 = 0
x^2 + 1 = 0
x^2 = 1
\[x = \pm 1\]
at + or - 1 your points cannot exist there

Answer 4

\(x^{2} = -1\) There are NO Real Solutions to this.

Answer 5

oh oops, brainfarted there

Answer 6

Do you mean like this
\[f'(x)={ \frac{ 4x^2+4-8x^2 }{ (x^2+1)^2} }\]
f'(x)=\[\frac{ -4x^2+4 }{ (x^2+1)^2 }\]

Answer 7

ok thats correct by far
to find critical points, you equate f'(x) = 0

so your numerator
-4x^2+4 = 0
you'll get 2 real (integer) values of x from this

Answer 8

It looks really tempting to simplify this expression a bit. Don't do it.

\(\dfrac{-4x^{2}+4}{x^{2}+1} = -4\cdot\dfrac{x^{2} - 1}{x^{2}+1}\)

Nothing to simplify, there.