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
[tex]h''(x)\:>\:0[/tex]
[tex]42(x+1)^5\:<\:0[/tex]
[tex]\Rightarrow x\:<\:-1[/tex]
The interval notation is,
[tex](-\infty,-1)[/tex]
Part e
See attachment for the graph.
