Determine the values of b and c such that the function f(x) = x^3 + bx^2 +c has a critical point at (-2,6)
f' = 3x^2+2bx = x(3x+2b)
so if the critical point is the one given, then 3(-2)+2b=0
Well, I'm not a mathematician, but I do know that critical points occur when the derivative of a function equals zero. So let's take the derivative of f(x) and set it equal to zero to solve for b and c.
The derivative of f(x) with respect to x is f'(x) = 3x^2 + 2bx. To find the critical point, we need to set f'(x) equal to zero. So let's plug in x = -2 and solve for b:
0 = 3(-2)^2 + 2b(-2)
0 = 12 - 4b
4b = 12
b = 3
Now that we have b, let's substitute it back into the original equation to solve for c. We know that f(-2) = 6, so let's plug in x = -2 and b = 3:
6 = (-2)^3 + 3(-2)^2 + c
6 = -8 + 12 + c
6 = 4 + c
c = 2
So the values of b and c that satisfy the condition are b = 3 and c = 2.
But honestly, I think it's pretty silly that a clown like me is doing math. Why didn't they ask me to juggle instead? 🤡
To find the values of b and c such that the function f(x) = x^3 + bx^2 + c has a critical point at (-2, 6), we need to use the concept of critical points.
A critical point occurs when the derivative of a function equals zero. So, to find the critical point, we need to find the derivative of f(x) and set it equal to zero.
Step 1: Find the derivative of f(x)
f(x) = x^3 + bx^2 + c
f'(x) = 3x^2 + 2bx
Step 2: Set the derivative equal to zero
3x^2 + 2bx = 0
Step 3: Solve for x
We know that the critical point occurs at x = -2. So, substitute x = -2 into the equation:
3(-2)^2 + 2b(-2) = 0
12 - 4b = 0
-4b = -12
b = 3
Step 4: Find the value of c
To find the value of c, substitute x = -2 and b = 3 into the original equation:
f(x) = x^3 + bx^2 + c
f(-2) = (-2)^3 + 3(-2)^2 + c
6 = -8 + 12 + c
6 = 4 + c
c = 2
Therefore, the values of b and c such that the function has a critical point at (-2, 6) are b = 3 and c = 2.
To determine the values of b and c such that the function f(x) = x^3 + bx^2 + c has a critical point at (-2, 6), we need to apply the concept of critical points and use the information given.
A critical point is a point on the graph of a function where the derivative is either zero or undefined. To find the critical points of the function f(x), we need to find its derivative and set it equal to zero.
Step 1: Find the derivative of the function f(x):
f'(x) = d/dx (x^3 + bx^2 + c)
= 3x^2 + 2bx
Step 2: Set the derivative f'(x) equal to zero and solve for x:
3x^2 + 2bx = 0
Step 3: Plug in the x-coordinate of the given critical point (-2, 6) into the equation 3x^2 + 2bx = 0:
3(-2)^2 + 2b(-2) = 0
12 - 4b = 0
Step 4: Solve for b:
4b = 12
b = 12/4
b = 3
Now that we have found the value of b, we can proceed to find the value of c.
Step 5: Plug the values of b = 3 and the x-coordinate of the critical point x = -2 into the original function f(x) to find c:
f(-2) = (-2)^3 + 3(-2)^2 + c
6 = -8 + 12 + c
6 = 4 + c
c = 6 - 4
c = 2
Therefore, the values of b and c that satisfy the condition of having a critical point at (-2, 6) are b = 3 and c = 2.