Consider the function below. (If an answer does not exist, enter DNE.) h(x) = (x + 1)7 − 7x − 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.) (d) Use the information from parts (a)-(c) to sketch the graph. Check your work with a graphing device if you have one.
Accepted Solution
A:
Part ai)
The given function is:
[tex]h(x)=(x+1)^7-7x-3[/tex]
The interval of increase is given by:
[tex]h'(x)\:>\:0[/tex]
[tex]\Rightarrow 7(x+1)^6-7\:>\:0[/tex]
[tex]x\:<\:-2,x\:>\:0[/tex]
In interval notation, we have;
[tex](-\infty,-2)\cup (0,+\infty)[/tex]
Part aii)
The interval of decrease is given by:
[tex]h'(x)\:<\:0[/tex]
[tex]\Rightarrow 7(x+1)^6-7\:>\:0[/tex]
[tex]\Rightarrow -2\:<\:x\:<\:0[/tex]
In interval notation, we have;
[tex](-2,0)[/tex]
Part bi)
At stationery points,
[tex]h'(x)=0[/tex]
[tex]\Rightarrow 7(x+1)^6-7=0[/tex]
[tex]\Rightarrow (x+1)^6=1[/tex]
[tex]\Rightarrow x+1=\pm1[/tex]
[tex]\Rightarrow x=-1\pm1[/tex]
[tex]\Rightarrow x=-2,x=0[/tex]
The stationary points are (0,-2) and (-2,10)
We use the second derivative test.
[tex]h"(x)=42(x+1)^5[/tex]
At local minimum,[tex]h'(x)=0[/tex] and [tex]h''(x)\:>\:0[/tex]
[tex]\Rightarrow h''(0)=42\:>\:0[/tex].
Hence the local minimum is (0,-2) it occurs at x=0.
At local maximum,[tex]h'(x)=0[/tex] and [tex]h''(x)\:<\:0[/tex]
[tex]\Rightarrow h''(-2)=-42\:<\:0[/tex].
The local maximum occurs at x=-2
Part C
At point of inflection,
[tex]h'(x)=0[/tex] and [tex]h''(x)\:=\:0[/tex]
There is no such stationary point.
DNE
Part di).
The graph is concave upwards for values where
[tex]h''(x)\:>\:0[/tex]
[tex]42(x+1)^5\:>\:0[/tex]
[tex]\Rightarrow x\:>\:-1[/tex]
In interval notation we have,
[tex](-1,+\infty)[/tex]
Part d ii) The graph is concave downwards for values where