Question

Find critical point(s) and intervals of concavity of function f(x) = (x -4)^2/3

Answer 1

Solution

Step 1

$\begin{gathered} \mathrm{Critical\:points\:are\:points\:where\:the\:function\:is\:defined\:and\:} \\ \begin{equation*} \mathrm{its\:derivative\:is\:zero\:or\:undefined} \end{equation*} \end{gathered}$

Step 2

$\begin{gathered} f(x)=\frac{\left(x-4\right)^2}{3} \\ \\ f^{\prime}(x)=\frac{2}{3}\left(x-4\right) \end{gathered}$

Step 3

$\begin{gathered} \frac{2}{3}\left(x-4\right)=0 \\ \\ x-4=0 \\ \\ x\text{ = 4} \end{gathered}$

Critical point is x = 4

Step 4

Concavity intervals definition

$\begin{gathered} \mathrm{If}\:f\:''\left(x\right)>0\:\mathrm{then}\:f\left(x\right)\:\mathrm{concave\:upwards.} \\ \\ \mathrm{If}\:f\:''\left(x\right)<0\:\mathrm{then}\:f\left(x\right)\:\mathrm{concave\:downwards.} \\ \mathrm{If}\:f\:^{\doubleprime}\left(x\right)\text{ = }\frac{2}{3} \end{gathered}$

Step 5

[tex]\mathrm{Concave\:Upward}:-\infty \:Final answer

[tex]\begin{gathered} Critical\text{ }pointis\text{ }x=4 \\ \mathrm{Concave\:Upward}\text{ on interval:}-\infty\: