increase/decrease & concavity

[startingover]

I think I have worked the problem correctly, but I am still uncomfortable with concavity and increase/decrease. Would you double check me please?

f(x) = (x^2-3)^2
f'(x) = 4x(x^2-3) or 4x^3 - 12x
f"(x) = 12x^2 - 12

critical first order numbers: square root 3,0
critical second order numbers: 0,9 & square root 1,0

increasing x<0, x>square root 3
decreasing 0<x<square root 3

concave up x<-1, x>1
concave down -1>x>1

 

[tkhunny]

You seem to be overlooking the third solution to f'(x) = 0. You managed the positive square root. Where is the negative square root?

 

[soroban]

Hello, startingover!

tkhunny is right . . . your critical values are off.


We are given: \(\displaystyle \:f(x) \:= \x^2\,-\,3)^2\)

The first derivative is: \(\displaystyle \:f'(x) \:= \:4x(x^2\,-\,3)\)
Solving: \(\displaystyle \:4x(x^2\,-\,3)\:=\:0\), we get: \(\displaystyle \:x\,=\,0,\,\pm\sqrt{3}\)
. . . . . three critical points.

\(\displaystyle - - - + - - - + - - - + - - -\)
. . . .-\(\displaystyle \sqrt{3}\;\;\;\;0\;\;\;\;\sqrt{3}\)

Test the slope in each of the four intervals.

\(\displaystyle \begin{array}{cccc}\text{On }(-\infty,\,-\sqrt{3})\text{, try }x\,=\,-2: &\:f'(-2)\:=\:-8(4 - 3)\:=\:-8\: & \text{decreasing: }\searrow \\
\text{On }(-\sqrt{3},\,0)\text{], try }x\,=\,-1: & \:f'(-1)\:=\:-4(1-3) \:=\:+12\: & \text{increasing: }\nearrow \\
\text{On }(0,\,\sqrt{3})\text{, try }x\,=\,1: &\:f'(1)\:=\:4(1-3)\:=\:-12\: & \text{decreasing: }\searrow \\
\text{On }(\sqrt{3},\,\infty)\text{, try }x\,=\,2: & \:f'(2)\:=\:8(4-3)\:=\:+8\: & \text{increasing: }\nearrow \end{array}\)

Therefore: \(\displaystyle \:\fbox{\begin{array}{cccc}\text{increasing:} & (-\sqrt{3}\,0) & \,\cup\, & (\sqrt{3},\,\infty) \\
\text{decreasing:} & \-\infty,\,-\sqrt{3}) & \cup & (0,\,\sqrt{3})
\end{array}}\)


The second derivative is: \(\displaystyle \:f''(x)\:=\:12x^2\,-\,12\)
Solving: \(\displaystyle \:12(x^2\,-\,1) \:=\:0\), we get: \(\displaystyle \:x\,=\,\pm1\)
. . . . . two inflection points.

\(\displaystyle - - - + - - - - + - - -\)
. . . .-\(\displaystyle 1\;\;\;\;\;\;1\)

Test the concavity in each of the three intervals.

\(\displaystyle \begin{array}{cccc}\text{On }(-\infty,\,-1)\text{, try }x\,=\,-2: &\:f''(-2)\:=\:12(4-1)\:=\:+36\: & \text{concave up: }\cup \\
\text{On }(-1,\,1)\text{, try }x\,=\,0: & \:f''(0)\:=\:12(0-1) \:=\:-12\: & \text{concave down: }\cap \\
\text{On }(1,\,\infty)\text{, try }x\,=\,2: & \:f''(2)\:=\:12(4-1)\:=\:+36\: & \text{concave up: }\cup\end{array}\)

Therefore: \(\displaystyle \:\fbox{\begin{array}{cc}\text{concave up: } & \-\infty,\,-1)\,\cup\,(1,\infty) \\
\text{concave down: } & \-1,\,1) \end{array}}\)