1. ## [SOLVED] Open intervals

On the function$\displaystyle \frac{x^2}{x^2+4}$
$\displaystyle y' = \frac{8x}{(x^2+4)^2} y''=\frac{-24x^2+32}{(x^2+4)^3)}$
I thought the critical numbers would be taken from the 2nd derivative, but my teacher said from the 1st. Is it always like this? and how would i know to take it from the denominator or numerator?

Also into which equation would I plug the critical numbers into? The 1st deriv. or 2nd?

On the function$\displaystyle \frac{x^2}{x^2+4}$
$\displaystyle y' = \frac{8x}{(x^2+4)^2} y''=\frac{-24x^2+32}{(x^2+4)^3)}$
I thought the critical numbers would be taken from the 2nd derivative, but my teacher said from the 1st. Is it always like this? and how would i know to take it from the denominator or numerator?

Also into which equation would I plug the critical numbers into? The 1st deriv. or 2nd?
critical values are where a derivative is equal to 0 or is undefined.

any derivative can have critical values ... it just depends on what you want to accomplish.

for the 1st derivative ...

if a function f(x) has extrema, then they are located at critical values of f'(x)

for the second derivative ...

if a function f(x) has inflection points, then they are located at critical values of f''(x)

note that the converse to both of the above statements is not necessarily true.

for your function $\displaystyle y = \frac{x^2}{x^2+4}$ ...

$\displaystyle y' = \frac{8x}{(x^2+4)^2}$ is equal to 0 at x = 0 (just set the numerator equal to 0)

y' is defined for all x.

so, there is only one critical value for y' ... x = 0 , and using the 1st derivative test for extrema, the original function has a minimum at x = 0.

$\displaystyle y'' = \frac{-8(3x^2-4)}{(x^2+4)^3}$

y'' = 0 at $\displaystyle x = \pm \frac{2}{\sqrt{3}}$

y'' is defined for all x.

y'' has critical values at $\displaystyle x = \pm \frac{2}{\sqrt{3}}$

the original function has inflection points at both critical values since y'' changes sign at each value.