Question

The function f(x) = x(x+2)^3 has the first derivative f'(x) = (4x+2)(x+2)^2 and second derivative f"(x) = 12(x+1)(x+2).

A) Find the intervals where f(x) is increasing or decreasing.
B) Find the values of x where f(x) has local maximum and local minimum values (if there are any).
C) Find the intervals of concavity and the (x,y) coordinates of inflection points (if there are any).
D) Use the info from parts a - c to sketch the graph.

Answer 1

A) Increasing: (-0.5, ∞)

Decreasing: (-∞, -2) ∪ (-2, -0.5)

B) x = -¹/₂

C) Concave up: (-∞, -2) ∪ (-1, ∞)

Concave down: (-2, -1)

Points of inflection: (-2, 0) and (-1, -1)

D) See attachments.

Explanation:

Given function and derivatives:

`f(x) = x(x+2)^3`

`f'(x) = (4x+2)(x+2)^2`

`f''(x) = 12(x+1)(x+2)`

Part A

`\textsf{A function is \textbf{increasing} when the \underline{gradient is positive}}\implies f'(x) > 0`

`\textsf{A function is \textbf{decreasing} when the \underline{gradient is negative}} \implies f'(x) < 0`

Increasing function

`\begin{aligned}f'(x) & > 0\\(4x+2)(x+2)^2 & > 0\\2(2x+1)(x+2)^2 & > 0\\(2x+1)(x+2)^2 & > 0\end{aligned}`

Therefore, the function is increasing on the Interval:

`\left(-(1)/(2), \infty \right)`

Decreasing function

`\begin{aligned}f'(x) & < 0\\(4x+2)(x+2)^2 & < 0\\2(2x+1)(x+2)^2 & < 0\\(2x+1)(x+2)^2 & < 0\end{aligned}`

Therefore, the function is decreasing on the Interval:

`(- \infty, -2) \cup \left(-2,-(1)/(2) \right)`

Part B

Local minimum/maximum points occur when f'(x) = 0.

`\begin{aligned}f'(x) &= 0\\(4x+2)(x+2)^2 & = 0\\2(2x+1)(x+2)^2 & = 0\\(2x+1)(x+2)^2 &= 0\\\\2x+1&=0 \implies x=-(1)/(2)\\(x+2)^2&=0 \implies x=-2\end{aligned}`

`f''\left(-(1)/(2)\right) = 12\left(-(1)/(2)+1 \right)\left(-(1)/(2)+2 \right)=9 > 0\implies \textsf{minimum}`

`f''\left(-2\right) = 12\left(-2+1 \right)\left(-2+2 \right)=0\implies \textsf{point of inflection}`

Therefore, the value of x where f(x) has a local minimum is x = -¹/₂.

Part C

At a point of inflection, f''(x) = 0.

`\begin{aligned}f''(x) &=0\\12(x+1)(x+2)&=0\\(x+1)(x+2)&=0\\\\x+1&=0 \implies x=-1\\x+2&=0 \implies x=-2\end{aligned}`

Substitute the found values of x into the original function to the find the y-coordinates of the points of inflection:

`\implies f(-1) = -1(-1+2)^3=-1`

`\implies f(-2) = -2(-2+2)^3=0`

Therefore, the inflection points are:

• (-2, 0) and (-1, -1)

A curve y = f(x) is concave up if f''(x) > 0 for all values of x.

A curve y = f(x) is concave down if f''(x) < 0 for all values of x.

Concave up

`\implies f''(x) > 0`

`\implies 12(x+1)(x+2) > 0`

`\implies (x+1)(x+2) > 0`

Therefore, the function is concave up at:

`\left(-\infty,-2\right) \cup \left(-1,\infty \right)`

Concave down

`\implies f''(x) < 0`

`\implies 12(x+1)(x+2) < 0`

`\implies (x+1)(x+2) < 0`

Therefore, the function is concave down at:

`(-2,-1)`

Part D

See attachment 1 for how to sketch the graph using the information from parts A-C.

See attachment 2 for the final sketch of the graph.