# find intervals where the function is concave upward and concave downward. f(x)=(x-6)(x+2)^3x-6=0 => x=6 (x+2)^3 => 3(x+2)^2 => 6(x+2) => 6x+12=0  => 6x=-12 => x=-2

sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use the second order derivative to tell where the graph of function is concave up or down.

You need to find the first order derivative using the product rule such that:

`f'(x) = (x-6)'(x+2)^3 + (x-6)((x+2)^3)` '

`f'(x) = (x+2)^3 + 3(x+2)^2(x-6)`

You need to find the second order derivative such that:

`f''(x) = 3(x+2)^2 + 6(x + 2)(x - 6) + 3(x+2)^2`

`f''(x) = 6(x+2)^2 + 6(x + 2)(x - 6) `

You need to factor out  `6(x + 2)`  such that:

`f''(x) =6(x + 2)(x + 2 + x - 6) => f''(x) = 6(x + 2)(2x - 4)`

You need to find the inflection points such that:

`f''(x) = 0 => 6(x + 2)(2x - 4) = 0 => (x + 2)(2x - 4) = 0`

`x + 2 = 0 => x = -2`

`2x -4 = 0 => 2x = 4 => x = 2`

Notice that the values of f''(x) are negative for `x in (-2,2)`  and they are positive for `x in (-oo,-2)U(2,oo).`

Hence, the function is concave up over `(-oo,-2)U(2,oo)`  and it is concave down if `x in (-2,2).`