Question

How do you find the x coordinates of all points of inflection, final all discontinuities, and find the open intervals of concavity for `y=(2x+3)^2(x+1)^2` for `[-10,0]`?

Answer 1

Explanation below
graph{(2x+3)^2(x+1)^2 [-2.035, 0.098, -0.503, 0.5646]}

Explanation:

The domain: `[-10,0]`

`f^'(x)=4(2x+3)(x+1)^2+2(2x+3)^2(x+1)`

`f^'(x)=2(2x+3)(x+1)[2(x+1)+2x+3]=2(2x+3)(x+1)(4x+5)`

`2x+3=0quad^^quadx+1=0quad^^quad4x+5=0`
`x=-3/2quad^^quadx=-1quad^^quadx=-5/4`

`x in [-10,-3/2] hArr f^'<0 =>`f goes down
`x=-3/2 =>` minimum `f(-3/2)=0`
`x in [-3/2,-5/4] hArr f^'>0 =>`f goes up
`x=-5/4 =>` local maximum
`x in [-5/4,-1] hArr f^'<0 =>`f goes down
`x=-1 =>` minimum `f(-1)=0`
`x in [-1,0] hArr f^'>0 =>`f goes up

Global maximum:
`f(-10)=(-20+3)^2(-10+1)^2=23409`

Cocavity:
`f^'(x)=4(2x+3)(x^2+2x+1)+2(4x^2+12x+9)(x+1)`

`f^'(x)=2(8x^3+30x^2+37x+15)`

`f^''(x)=2(24x^2+60x+37)`

`x_(1,2)=(-b+-sqrt(b^2-4ac))/(2a)=(-60+-sqrt(60^2-4*24*37))/(2*24)`

`x_(1,2)=(-15+-sqrt(3))/(12)`

`x_(1)=-1.394...`
`x_(2)=-1.105...`

`x in [-10,x_1) hArr f^''>0 =>` Convex
`x in (x_1,x_2) hArr f^''<0 =>` Concave
`x in (x_2,0] hArr f^''>0 =>` Convex