Question

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

Answer 1

Step 1

Step 2

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

Step 3

`\begin{gathered} (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{ = }(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\: