Find a cubic polynomial such as f(x)=ax^3+bx^2+cx+d
that has horizontal tangents at the points (-5,4) and (4,-4)
clearly, we want
f'(x) = 3ax^2 + 2bx + c
to be zero at x= -5 and 4. So
3ax^2 + 2bx + c = k(x+5)(x-4) = kx^2+kx-20k
so, c=-20k, 2b=k, 3a=k and
f(x) = (k/3)x^3 + (k/2)x^2 - 20kx + d
f(4) = 64k/3 + 8k - 80k + d = -152k/3 + d = -4
f(-5) = -125k/3 + 25k/2 + 100k + d = 425k/6 + d = 4
solve those two and you get
k = 48/729
d = -484/729
f(x) = 16/729 x^3 + 24/729 x^2 - 960/729 x + -484/729
or,
1/729 (16x^3 + 24x^2 - 960x - 484)
You can see the curve at
http://www.wolframalpha.com/input/?i=16%2F729+x^3+%2B+24%2F729+x^2+-+960%2F729+x+%2B+-484%2F729
scroll down to where it lists the relative max,min to verify that it meets the requirements.
To find a cubic polynomial that has horizontal tangents at the given points, we need to use some key facts about tangent lines.
First, we know that a horizontal line has a slope of zero. Therefore, the tangent line at any point on the graph of the cubic polynomial will have a slope of zero if it is a horizontal tangent.
Second, we can use the fact that the slope of the tangent line at a point is equal to the derivative of the function at that point.
Let's start by finding the derivative of the cubic polynomial function f(x) = ax^3 + bx^2 + cx + d.
Step 1: Compute the derivative of f(x)
To find the derivative of a polynomial function, we differentiate each term individually. The derivative of xn is given by nx^(n-1).
f'(x) = 3ax^2 + 2bx + c
Step 2: Set the derivative equal to zero for each given point
Since the tangent lines at (-5,4) and (4,-4) have slopes of zero, the derivative of f(x) must be equal to zero at these points.
For the point (-5,4):
0 = 3a(-5)^2 + 2b(-5) + c
Simplifying and rearranging, we get:
25a - 10b + c = 0 ----(Equation 1)
For the point (4,-4):
0 = 3a(4)^2 + 2b(4) + c
Simplifying and rearranging, we get:
32a + 8b + c = 0 ----(Equation 2)
Step 3: Solve the system of equations
To find the values of a, b, and c that satisfy both Equation 1 and Equation 2, we need to solve this system of equations.
Subtracting Equation 1 from Equation 2, we eliminate the c term:
32a + 8b + c - (25a - 10b + c) = 0
Simplifying, we have:
32a + 8b - 25a + 10b = 0
7a + 18b = 0
We have one equation with two variables. To solve for a and b, we need an additional equation. Since this is a cubic polynomial, we need one more piece of information.
Let's assume that the cubic polynomial has a minimum or maximum point at the origin (0,0). This assumption satisfies the given conditions.
Step 4: Set the value of the polynomial at x = 0 to 0
We have the function f(x) = ax^3 + bx^2 + cx + d. Plugging in x = 0 and y = 0, we have:
0 = a(0)^3 + b(0)^2 + c(0) + d
0 = d
Therefore, d = 0.
Step 5: Now, we have the values of d = 0 and the equation 7a + 18b = 0.
Substitute d = 0 into the equation, we get:
7a + 18b = 0
Solve the system of equations:
7a + 18b = 0 ----(Equation 3)
7a = -18b
Divide both sides of Equation 3 by 7:
a = -18b/7
Step 6: Substitute the value of a into Equation 1 or 2 to solve for b.
Using Equation 1, we substitute a = -18b/7:
25(-18b/7) - 10b + c = 0
Multiply through by 7 to clear the fraction:
-450b - 70b + 7c = 0
Combine like terms:
-520b + 7c = 0 ----(Equation 4)
Step 7: Solve for b and c
We now have the system of equations: Equation 3 (-18a/7 = b) and Equation 4 (-520b + 7c = 0).
Substituting the value of a from Equation 3 into Equation 4, we get:
-520(-18b/7) + 7c = 0
Simplify and solve for b:
9360b/7 + 7c = 0
9360b + 49c = 0
9360b = -49c
Divide both sides by 49:
190.2b = -c
We can choose any non-zero value for b. Let's use b = 1 for simplicity.
Substitute b = 1 into the equation:
190.2(1) = -c
c = -190.2
Now substitute the values of b = 1 and c = -190.2 into Equation 3 to solve for a:
7a + 18(1) = 0
7a + 18 = 0
7a = -18
a = -18/7
Therefore, the cubic polynomial is:
f(x) = (-18/7)x^3 + x^2 - 190.2x
Note that we assume d = 0 since we require a cubic polynomial with horizontal tangents at the given points.