Question

consider the following function. (if an answer does not exist, enter dne.)
h(x)=(x + 1)^5-5x - 3
(a) find the interval of increase. (enter your answer using interval notation.)
find the interval of decrease. (enter your answer using interval notation.)
(b) find the local minimum value(s). (enter your answers as a comma - separated list.)
find the local maximum value(s). (enter your answers as a comma - separated list.)
(c) find the inflection point.
(x,y)=()
find the interval where the graph is concave upward. (enter your answer using interval notation.)
find the interval where the graph is concave downward. (enter your answer using interval notation.)

Explanation:

Step1: Find the first derivative

Using the chain rule \((u^n)^\prime = nu^{n - 1}u^\prime\) and \((x^n)^\prime=nx^{n - 1}\), for \(h(x)=(x + 1)^5-5x - 3\), let \(u=x + 1\), then \(h^\prime(x)=5(x + 1)^4-5=5[(x + 1)^4-1]=5[(x + 1)^2+1][(x + 1)^2 - 1]=5[(x + 1)^2+1](x+2)x\).
Set \(h^\prime(x)=0\), we get \(x=-2\) and \(x = 0\) (since \((x + 1)^2+1>0\) for all real \(x\)).

Step2: Determine intervals of increase and decrease

We consider the intervals \((-\infty,-2)\), \((-2,0)\) and \((0,\infty)\).

• For \(x\in(-\infty,-2)\), let \(x=-3\), then \(h^\prime(-3)=5[(-3 + 1)^2+1](-3 + 2)(-3)=5\times(4 + 1)\times(-1)\times(-3)=75>0\).
• For \(x\in(-2,0)\), let \(x=-1\), then \(h^\prime(-1)=5[(-1 + 1)^2+1](-1 + 2)(-1)=5\times(0 + 1)\times1\times(-1)=-5<0\).
• For \(x\in(0,\infty)\), let \(x = 1\), then \(h^\prime(1)=5[(1 + 1)^2+1](1 + 2)(1)=5\times(4+1)\times3\times1 = 75>0\).

So the function \(h(x)\) is increasing on \((-\infty,-2)\cup(0,\infty)\) and decreasing on \((-2,0)\).

Step3: Find local extrema

Since \(h(x)\) changes from increasing to decreasing at \(x=-2\), \(h(-2)=(-2 + 1)^5-5\times(-2)-3=-1 + 10-3 = 6\).
Since \(h(x)\) changes from decreasing to increasing at \(x = 0\), \(h(0)=(0 + 1)^5-5\times0-3=1-3=-2\).

Step4: Find the second derivative

\(h^\prime(x)=5(x + 1)^4-5\), then \(h^{\prime\prime}(x)=20(x + 1)^3\).
Set \(h^{\prime\prime}(x)=0\), we get \(x=-1\).
When \(x=-1\), \(h(-1)=(-1 + 1)^5-5\times(-1)-3=0 + 5-3=2\).
For \(x\in(-\infty,-1)\), let \(x=-2\), \(h^{\prime\prime}(-2)=20(-2 + 1)^3=-20<0\).
For \(x\in(-1,\infty)\), let \(x=0\), \(h^{\prime\prime}(0)=20(0 + 1)^3=20>0\).

Answer:

(a)

• Interval of increase: \((-\infty,-2)\cup(0,\infty)\)
• Interval of decrease: \((-2,0)\)

(b)

• Local minimum value: \(-2\)
• Local maximum value: \(6\)

(c)

• Inflection point: \((-1,2)\)
• Concave upward: \((-1,\infty)\)
• Concave downward: \((-\infty,-1)\)