# Determine the values of the constants α and β so that the function f (x) = x3 + αx2 + βx + δ may have a relative maximum at x = −3, and a relative minimum at x = 1.

## For ease of typing, I will use

f(x) = x^3 + ax^2 + bx + c

take the derivative and the c will drop out
set this equal to zero, with x = -3 to get an equation in a and b
repeat with x = 1

you now have two equations in a and b, use your favourite grade 9 method to solve them

## To determine the values of the constants α and β, we will need to use the properties of the first and second derivatives of the function.

Step 1: Find the first derivative of the function f(x):
f'(x) = 3x^2 + 2αx + β

Step 2: Find the second derivative of the function f(x):
f''(x) = 6x + 2α

Step 3: Determine the conditions for a relative maximum at x = -3 and a relative minimum at x = 1.

At x = -3:
For a relative maximum at x = -3, we need f'(x) = 0 and f''(x) < 0.

Substitute x = -3 into f'(x) and solve for α:
3(-3)^2 + 2α(-3) + β = 0
9α - 6α + β = 0
3α + β = 0

Substitute x = -3 into f''(x) and solve for α:
6(-3) + 2α < 0
-18 + 2α < 0
2α < 18
α < 9

At x = 1:
For a relative minimum at x = 1, we need f'(x) = 0 and f''(x) > 0.

Substitute x = 1 into f'(x) and solve for α:
3(1)^2 + 2α(1) + β = 0
3 + 2α + β = 0

Substitute x = 1 into f''(x) and solve for α:
6(1) + 2α > 0
6 + 2α > 0
2α > -6
α > -3

Step 4: Combine the conditions and solve for α and β.
From the condition at x = -3, we have α < 9. From the condition at x = 1, we have α > -3.

Therefore, the values of α must satisfy -3 < α < 9.

Since the question does not provide a specific value for δ, we cannot determine the exact values of α and β.

## To determine the values of the constants α and β, we can use the properties of a relative maximum and a relative minimum.

A relative maximum occurs at a point where the derivative of the function changes from positive to negative. Conversely, a relative minimum occurs at a point where the derivative changes from negative to positive.

First, let's find the derivative of the function f(x):

f(x) = x^3 + αx^2 + βx + δ

f'(x) = 3x^2 + 2αx + β

To have a relative maximum at x = -3, the derivative f'(x) should change sign from positive to negative at x = -3. This means that f'(-3) should be greater than 0, and f'(-3+h) should be less than 0 for some small positive value of h.

Using the derivative f'(x), we can evaluate this condition:

f'(-3) = 3(-3)^2 + 2α(-3) + β
= 9 - 6α + β

To have a relative minimum at x = 1, the derivative f'(x) should change sign from negative to positive at x = 1. This means that f'(1) should be less than 0, and f'(1+h) should be greater than 0 for some small positive value of h.

Using the derivative f'(x), we can evaluate this condition:

f'(1) = 3(1)^2 + 2α(1) + β
= 3 + 2α + β

Now we have two equations that we can use to solve for α and β:

Condition 1: 9 - 6α + β > 0
Condition 2: 3 + 2α + β < 0

Solving these two inequalities simultaneously can determine the values of α and β that satisfy both conditions and allow for a relative maximum at x = -3 and a relative minimum at x = 1.