consider the following function. (if an answer does not exist, enter dne.)\nh(x)=(x + 1)^5-5x - 3\n(a) find…

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+1x). 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))