Question

Find the open intervals on which the function is concave up or concave down
f(x)=1/ x^2 + 3 WILL GIVE MEDAL PLEASE HELP!

Answer 1

what did you get for the first derivative?

Answer 2

X^2+3-2x/(x^2+3)^2

Answer 3

Where did you go @satellite73 ?

Answer 4

Hmm I'm a little worried about your first derivative.
Let's simply use the power rule to do this one.\[\large f(x)=\frac{1}{x^2+3} \qquad=\qquad (x^2+3)^{-1}\]

Taking the derivative gives us,\[\large f'(x)=-(x^2+3)^{-2}(2x) \qquad=\qquad \frac{-2x}{(x^2+3)^2}\]

Answer 5

Did you apply the quotient rule to do this one?

Answer 6

Yeah so I guess that's why it was off

Answer 7

Quotient Rule will give you the same result.
Lemme show you where the mistake was.

Answer 8

\[\large f'(x)=\frac{\color{royalblue}{(1)'}(x^2+3)-(1)\color{royalblue}{(x^2+3)'}}{(x^2+3)^2}\]

Answer 9

So there is our setup for the quotient rule.

Answer 10

What's the derivative of 1?
Derivative of a constant, hmmm.

Answer 11

LOL oops! It's been a long day and I'm just so stressed :/

Answer 12

:3

Answer 13

So now what do I do once I find the derivative?

Answer 14

\[\large f'(x)=0 \qquad\to\qquad \text{critical points}\]When we look at test points around the critical points, it tells us where the function is `increasing and decreasing`.

\[\large f''(x)=0 \qquad\to\qquad \text{inflection points}\]When we look at test points around the inflection points, it tells us where the function is `concave up and concave down`.

So we'll want to take the derivative once more, then set it equal to zero.

Answer 15

So looks like we'll need to apply the Quotient Rule again.

Answer 16

So here is our setup,
\[\large f''(x)=\frac{\color{royalblue}{(-2x)'}(x^2+3)^2-(-2x)\color{royalblue}{\left[(x^2+3)^2\right]'}}{\left[(x^2+3)^2\right]^2}\]