Consider the function f(x)=x^3+ax^2+bx+c that has a relative minimum at x=3 and an inflection point at x=2.
a). Determine the constants a and b to make the above information true for this function.
b). Find a relative maximum point of the function found in part (a) after plugging in values for a and b, given that f(0)=5.
f' = 3x^2 + 2ax + b
f" = 6x+2a
You know that
f"(2) = 0
12+2a = 0
a = -6
Now do the same with f'(3)=0
a) To determine the constants a and b, we need to use the given information about the relative minimum and inflection point.
1. Relative minimum at x = 3:
The first derivative of the function will be zero at x = 3.
f'(x) = 3x^2 + 2ax + b
Setting x = 3 and solving for a and b:
0 = 3(3)^2 + 2a(3) + b
0 = 27 + 6a + b ...(1)
2. Inflection point at x = 2:
The second derivative of the function will be zero at x = 2.
f''(x) = 6x + 2a
Setting x = 2 and solving for a:
0 = 6(2) + 2a
0 = 12 + 2a
-2a = 12
a = -6 ...(2)
Plugging a = -6 into equation (1):
0 = 27 + 6(-6) + b
0 = 27 - 36 + b
0 = -9 + b
b = 9
Therefore, the constants a = -6 and b = 9.
b) To find the relative maximum point of the function, we need to find the x-coordinate of the maximum point and then substitute it into the function to find the corresponding y-coordinate.
1. Find x-coordinate:
The x-coordinate of the relative maximum point corresponds to another point where the first derivative is zero.
Setting f'(x) = 0 and solving for x:
3x^2 + 2ax + b = 0
Substituting the values of a = -6 and b = 9:
3x^2 - 12x + 9 = 0
Dividing by 3: x^2 - 4x + 3 = 0
Factoring: (x - 3)(x - 1) = 0
x = 3 or x = 1
Since x = 3 corresponds to the relative minimum, the x-coordinate of the relative maximum point is x = 1.
2. Find y-coordinate:
Substituting x = 1 into the function:
f(1) = (1)^3 + (-6)(1)^2 + 9(1) + c
f(1) = 1 - 6 + 9 + c
f(1) = 4 + c
Given f(0) = 5, we can substitute x = 0 into the function:
f(0) = (0)^3 + (-6)(0)^2 + 9(0) + c
f(0) = 0 + 0 + 0 + c
f(0) = c
So, c = 5.
Therefore, after plugging in values for a = -6, b = 9, and c = 5, the relative maximum point is (1, 4 + c) = (1, 9).
To determine the constants a and b in the function f(x) = x^3 + ax^2 + bx + c, let's use the information given:
a) Relative minimum at x = 3:
To find the relative minimum of a function, we need to locate the critical point, which occurs at the point where the derivative is zero. Since the function has a relative minimum at x = 3, we know that f'(3) = 0.
Taking the derivative of f(x) = x^3 + ax^2 + bx + c, we get:
f'(x) = 3x^2 + 2ax + b
Substituting x = 3 into the derivative equation:
f'(3) = 3(3)^2 + 2a(3) + b = 0
27 + 6a + b = 0
b) Inflection point at x = 2:
To find the inflection point of a function, we need to find the second derivative and check where it changes sign. Since the function has an inflection point at x = 2, we know that f''(2) = 0.
Taking the second derivative of f(x), we get:
f''(x) = 6x + 2a
Substituting x = 2 into the second derivative equation:
f''(2) = 6(2) + 2a = 0
12 + 2a = 0
Now we have two equations:
27 + 6a + b = 0 ---- (Equation 1)
12 + 2a = 0 ---- (Equation 2)
Solving Equation 2 for a:
2a = -12
a = -6
Substituting the value of a into Equation 1:
27 + 6(-6) + b = 0
27 - 36 + b = 0
b - 9 = 0
b = 9
Therefore, the constants a = -6 and b = 9 will make the function have a relative minimum at x = 3 and an inflection point at x = 2.
b) Finding the relative maximum point:
To find the relative maximum point, we need to find the critical point where the derivative changes sign. Since the function is cubic, it does not have a relative maximum point.
However, the question states that f(0) = 5. Plugging in x = 0 into the original equation:
f(0) = (0)^3 + a(0)^2 + b(0) + c = 0 + 0 + 0 + c = c
So, c = 5.
Therefore, the relative maximum point of the function when a = -6, b = 9, and c = 5 is not defined.