Find a cubic function
f(x) = ax3 + bx2 + cx + d that has a local maximum value of 3 at x = −4 and a local minimum value of 0 at x = 2.
f'(x) = 3ax^2 + 2bx + c = m(x+4)(x-2) = mx^2+2mx-8m
f"(x) = 6ax+2b = 2m(x+1)
since f"(-4) < 0 and f"(2) > 0, m > 0
but you already knew that, just from what you know about the shape of cubics.
Now, we have b = 3a and c = -8b and m=b, so
f(x) = b/3 x^3 + bx^2 - 8bx + d
But, we know that f(-4) = 3 and f(2) = 0, so
-64/3 b + 16b + 32b + d = 3
8/3 b + 4b - 16b + d = 0
b = 1/12, d = 7/9
which gives us
f(x) = 1/36 x^3 + 1/12 x^2 - 2/3 x + 7/9
= 1/36(x^2+3x^2-24x+28)
The desired properties can be found if you scroll down a bit at
www.wolframalpha.com/input/?i=1%2F36+x%5E3+%2B+1%2F12+x%5E2+-+2%2F3+x+%2B+7%2F9
y = a x^3 + b x^2 + c x + d
dy/dx = 3 a x^2 + b x + c is = 0 at x = -4 and at x = 2
d^2 y/dx^2 = 6 a x + b is negative at x = -4 and positive at x = 2
so at x = -4
-64 a + 16 b - 4 c+ d = 0
3 a (16) -4b +c = 0
and at x = 2
8 a +4 b + 2 c + d = 0
3 a (4)+ 2 b + c =0
find a, b and c and d
whoops, 3 not 0 at x = -4
so at x = -4
-64 a + 16 b - 4 c+ d = 3
Why did the cubic function go to the comedy club? It was looking for some local maximum and minimum values! Okay, okay, I'll stop clowning around and get to the point!
Let's find the cubic function that fits the given criteria. We know that at x = -4, the local maximum value is 3, so we have the point (-4, 3). Similarly, at x = 2, the local minimum value is 0, so we have the point (2, 0).
To find the values of a, b, c, and d, we can use these points.
Substituting the x and y values of the first point (-4, 3) into the cubic function, we get:
3 = a(-4)^3 + b(-4)^2 + c(-4) + d
Simplifying that equation, we obtain:
3 = -64a + 16b - 4c + d
Now, let's substitute the x and y values of the second point (2, 0) into the cubic function:
0 = a(2)^3 + b(2)^2 + c(2) + d
Simplifying this equation, we have:
0 = 8a + 4b + 2c + d
Now, we have two equations:
-64a + 16b - 4c + d = 3
8a + 4b + 2c + d = 0
Solving these equations simultaneously, we find:
a = 1/4
b = -2
c = -5/2
d = 5/2
Therefore, the cubic function that satisfies the given conditions is:
f(x) = (1/4)x^3 - 2x^2 - (5/2)x + 5/2
Now, that's a cubic function that knows how to have some local fun at specific x-values!
To find a cubic function with the given local maximum and minimum values, we can start by determining the critical points of the function.
The critical points occur where the derivative of the function is zero or undefined. In this case, we are given that there is a local maximum at x = -4 and a local minimum at x = 2.
Since the local maximum occurs at x = -4, the derivative of the function at that point is zero. Similarly, since the local minimum occurs at x = 2, the derivative of the function at that point is also zero.
Let's proceed step by step to find the function:
Step 1: Find the derivative of the cubic function:
f'(x) = 3ax^2 + 2bx + c
Step 2: Set up the equations using the given information:
f'(-4) = 0
=> 3a(-4)^2 + 2b(-4) + c = 0
f'(2) = 0
=> 3a(2)^2 + 2b(2) + c = 0
Step 3: Solve the system of equations:
Using the given information, we can set up the following equations:
16a - 8b + c = 0 ----(1)
12a + 4b + c = 0 ----(2)
Step 4: Solve the system of equations:
To solve the system of equations, we can eliminate the variable 'c'. Subtracting equation (2) from equation (1), we get:
4a - 12b = 0
Dividing both sides by 4, we have:
a - 3b = 0 ----(3)
From equation (3), we can express 'a' in terms of 'b' as:
a = 3b
Step 5: Substitute the value of 'a' into either equation (1) or (2):
Let's substitute 'a' in equation (1):
16a - 8b + c = 0
16(3b) - 8b + c = 0
48b - 8b + c = 0
40b + c = 0 ----(4)
Step 6: Substitute the value of 'a' and 'c' back into equation (2):
Using equation (3), we have:
12a + 4b + c = 0
12(3b) + 4b + c = 0
36b + 4b + c = 0
40b + c = 0 ----(5)
Comparing equations (4) and (5), we observe that they are the same. This suggests that the value of 'b' can be anything.
Step 7: Choose a value for 'b':
To determine the specific cubic function, we need to choose a value for 'b'. Let's assume 'b' as 1 for simplicity.
Step 8: Substitute the chosen value of 'b' into equations (3), (4), and (5):
a = 3b becomes:
a = 3(1) = 3
Substituting 'b' into equations (4) and (5), we get:
40(1) + c = 0
=> 40 + c = 0
=> c = -40
Step 9: Substitute the values of 'a', 'b', and 'c' into the original equation:
The cubic function f(x) = ax^3 + bx^2 + cx + d becomes:
f(x) = 3x^3 + x^2 - 40x + d
Step 10: Determine the value of 'd' by using one of the given points:
We know that f(x) has a local maximum of 3 at x = -4.
Substituting these values into the function, we have:
3 = 3(-4)^3 + (-4)^2 - 40(-4) + d
3 = -192 + 16 + 160 + d
3 = -16 + d
d = 3 + 16
d = 19
Thus, the cubic function that satisfies the given conditions is:
f(x) = 3x^3 + x^2 - 40x + 19.