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

giorgiana1976 | Student

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).

neela | Student

y(x)= x^4+32x+40.

To find the relative maximum.

WE find dy/dx and set it to zero and solve for x. The silutions gives the values of x at which t becomes extreme. Again we substitute these values in d2y/dx^2  and find whether d2/dx^2  is negative (for maximum) or positive for minimum.

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

y' = (4x^3 +32 ) .. Equating to zero, we get:

4x^3+32 = 0

x^3+8 = 0

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

Therefore x = -2 is the only real root, as the other equation has its discriminant 2^2 -4*4 is  negative  and so no real roots.

y"(x) = (x^3+8)'

y"(x) = 3x^2.

y"(-2) = 3(-2)^2 = 6 is positive for x = -2

Therefore y(-2) = (-2)^4+32(-2)+40

y(-2) = 16-64+40 .

y(-2) = -8 is the minimum at x = -2.