A function is defined by f(x) = ax^3 + bx^2 + cx + d. Find a, b, c, and d if f(x) has a point of inflection at (0, 2) and a local maximum at (2,6).
y = ax^3 + bx^2 + cx + d
(2,6) lies on it, so 6 = 8a + 4b + 2c + d
(0,2) lies on it, so 2 = 0+0+0+d ----> d = 2
then 8a + 4b + 2c = 4
4a + 2b + c = 2 **
y' = 3ax^2 + 2bx + c
since (2,6) is a max
when x = 2, y' = 0
3a(4) + 2b(2) + c = 0
12a + 4b + c = 0 ***
subtract ** from ***
8a + 2b = -2
4a + b = -1
y'' = 6ax + 2b
(0,2) is a point of inflection, so when x = 0, y'' = 0
6a(0) + 2b = 0
2b = 0
b = 0
then in 4a + b = -1, a = -1/4
in ** , 4a + 2b + c = 2
4(-1/4) + 0 + c = 2
c = 3
f(x) = (-1/4)x^3 + 3x + 2
check: https://www.wolframalpha.com/input/?i=plot+f%28x%29+%3D+%28-1%2F4%29x%5E3+%2B+3x+%2B+2
Yup!
To find the values of a, b, c, and d, we can use the information given about the point of inflection and local maximum.
1. Point of Inflection (0, 2):
At the point of inflection, the second derivative of the function equals zero. Let's start by finding the second derivative of f(x).
The first derivative of f(x) gives us:
f'(x) = 3ax^2 + 2bx + c
Differentiating f'(x), we get the second derivative:
f''(x) = 6ax + 2b
Since f''(x) equals zero at the point of inflection (0, 2), we can substitute x = 0 and f''(x) = 0:
6a(0) + 2b = 0
This simplifies to:
2b = 0
From this equation, we know that b = 0.
2. Local Maximum (2, 6):
At the local maximum, the first derivative of the function equals zero. Let's use this information to find the remaining values of a, c, and d.
Using the given local maximum point (2, 6), we substitute x = 2 and f'(x) = 0:
3a(2)^2 + 2(0)(2) + c = 0
12a + c = 0
From this equation, we know that c = -12a.
Now, let's substitute the values b = 0 and c = -12a into the original equation for f(x):
f(x) = ax^3 + bx^2 + cx + d
f(x) = ax^3 + 0x^2 - 12ax + d
f(x) = ax^3 - 12ax + d
We still need to find the values of a and d.
Substitute the point (2, 6) into f(x):
f(2) = 6
a(2)^3 - 12a(2) + d = 6
8a - 24a + d = 6
-16a + d = 6
Since we have two variables (a and d) and one equation, we need more information to find their exact values.
To find the values of a, b, c, and d, we can use the given information about the point of inflection and local maximum.
We know that the point of inflection is (0, 2), which means the second derivative of f(x) is zero at x = 0. Therefore, we need to take the second derivative of f(x) and set it equal to zero to find a relationship between a, b, and c.
First, let's calculate the first derivative of f(x) using the power rule:
f'(x) = 3ax^2 + 2bx + c
Now, let's calculate the second derivative of f(x) by taking the derivative of the first derivative:
f''(x) = 6ax + 2b
Since the second derivative of f(x) is zero at x = 0 (point of inflection), we can substitute x = 0 into f''(x) to get:
6a(0) + 2b = 0
2b = 0
b = 0
Therefore, we have found that b = 0.
Now, let's consider the local maximum at (2, 6). This means that the first derivative of f(x) is zero at x = 2, and the second derivative of f(x) is negative at x = 2.
Setting the first derivative of f(x) to zero, we can solve for a and c:
3a(2^2) + 2(0)(2) + c = 0
12a + c = 0 ---- Equation (1)
Next, let's evaluate the second derivative of f(x) at x = 2:
6a(2) + 2(0) < 0
12a < 0
a < 0
Now, using Equation (1), we can substitute a = -1 into the equation to find c:
12(-1) + c = 0
-12 + c = 0
c = 12
So, we have found that a = -1, b = 0, and c = 12.
To find the value of d, we can substitute these values into the original equation of f(x) and use the fact that f(x) has a local maximum at (2, 6):
f(2) = (-1)(2^3) + (0)(2^2) + (12)(2) + d = 6
Simplifying the equation, we get:
-8 + 24 + 12 + d = 6
28 + d = 6
d = 6 - 28
d = -22
Therefore, we have found that a = -1, b = 0, c = 12, and d = -22.