Investigate for relative maximum or minimum points and points of inflection for y = x^4 +32x + 40

The relative extreme points of a function could be
determined by calculating the roots of the first derivative of the
function.

dy/dx = d/dx(x^4 +32x +
40)

dy/dx = 4x^3 + 32

We'll
put dy/dx = 0.

4x^3 + 32 =
0

We'll divide by 4:

x^3 + 8 =
0

We'll write the sum of cubes using the
formula:

(a^3 + b^3) =
(a+b)(a^2-ab+b^2)

a = x and b =
2

x^3 + 8 = x^3 + 2^3

x^3 +
2^3 = (x+2)(x^2 - 2x + 4)

(x+2)(x^2 - 2x + 4) =
0

W'll set each factor as 0:

x
+ 2 = 0

x = -2

x^2 - 2x + 4
> 0 for any real x.

The local extreme
point is in x = -2.

f(-2) = (-2)^4 +32*(-2)
+ 40

f(-2) = 16 + 40 -
64

f(-2) =
-8

The inflection points could be found
by calculating the roots of the second derivative (if there are
any).

f"(x) =
[f'(x)]'

[f'(x)]' = (x^3 +
8)'

[f'(x)]' =
3x^2

3x^2>0 for any real x, except x = 0 when 3x^2 =
0.

The inflection point is in x =
0.

f(0) = 0^4 +32*0 + 40

f(0)
= 40

The inflection point is:
(0,40).