find the values of a and b if the function y=2x^3+ax^2+bx+36 has a local maximum when x=-4 and a local minimum when x=5
Sure, I'll try to find the values of a and b for you. Before I do that, I must say that math and humor are like clowns and tightrope walking – a bit of a balancing act. Let's proceed with caution!
To find the values of a and b, we need to consider the behavior of the function at the given points. At a local maximum or minimum, the derivative of the function should be equal to zero.
So let's take the derivative of the function y = 2x^3 + ax^2 + bx + 36:
dy/dx = 6x^2 + 2ax + b
Now, since we want a local maximum at x = -4, we can set the derivative to zero and solve for a and b:
0 = 6(-4)^2 + 2a(-4) + b
0 = 96 - 8a + b (Equation 1)
Similarly, for a local minimum at x = 5:
0 = 6(5)^2 + 2a(5) + b
0 = 150 + 10a + b (Equation 2)
Now, to solve these two equations simultaneously, we can subtract Equations 1 and 2:
(96 - 8a + b) - (150 + 10a + b) = 0
-54 - 18a = 0
-18a = 54
a = -3
Substituting this value of a into Equation 1:
96 - 8(-3) + b = 0
96 + 24 + b = 0
b = -120
So, the values of a and b that satisfy the given conditions are a = -3 and b = -120. Voila!
Remember, this clown is an expert in math humor but sometimes the actual calculations can be a bit tricky. Hope this helps!
To find the values of a and b, we need to determine the derivatives of the given function, set them equal to zero, and use the given conditions to solve for a and b.
Step 1: Find the derivative of the function y with respect to x.
y = 2x^3 + ax^2 + bx + 36
The first derivative is:
dy/dx = 6x² + 2ax + b
Step 2: Set the derivative equal to zero to find the critical points.
6x² + 2ax + b = 0
Step 3: Substitute the given x-values (-4 and 5) to create two equations.
At x = -4:
6(-4)² + 2a(-4) + b = 0
16a - 8a + b = 0
8a + b = 0 -----(Equation 1)
At x = 5:
6(5)² + 2a(5) + b = 0
50a + 10a + b = 0
60a + b = 0 -----(Equation 2)
Step 4: Solve the system of equations (Equation 1 and Equation 2).
Multiply Equation 1 by 6:
48a + 6b = 0 ----(Equation 3)
Subtract Equation 3 from Equation 2 to eliminate b:
(60a + b) - (48a + 6b) = 0
12a - 5b = 0 ----(Equation 4)
Step 5: Solve Equations 3 and 4 simultaneously.
Multiply Equation 4 by 6:
72a - 30b = 0 ----(Equation 5)
Multiply Equation 3 by 2:
96a + 12b = 0 ----(Equation 6)
Add Equation 5 and Equation 6 to eliminate b:
(72a - 30b) + (96a + 12b) = 0
168a - 18b = 0
Divide both sides by 6:
28a - 3b = 0 ----(Equation 7)
We can now set Equation 7 equal to Equation 4 and solve for a:
28a - 3b = 12a - 5b
16a = 2b
8a = b
Substituting 8a for b in Equation 7:
28a - 3(8a) = 0
28a - 24a = 0
4a = 0
a = 0
Step 6: Substitute the value of a back into Equation 1 to solve for b:
8a + b = 0
8(0) + b = 0
b = 0
Therefore, the values of a and b are both 0.
To find the values of a and b, we need to use the information given about the local maximum and minimum of the function.
First, let's find the derivative of the given function y = 2x^3 + ax^2 + bx + 36. The derivative will give us the equation for the slope of the function at different points.
Differentiating the function with respect to x:
dy/dx = 6x^2 + 2ax + b
To find the critical points (where the local maximum and minimum occur), we set the derivative equal to zero and solve for x:
6x^2 + 2ax + b = 0
Given that the local maximum occurs when x = -4, substitute this value into the equation:
6(-4)^2 + 2a(-4) + b = 0
96 - 8a + b = 0 -- (equation 1)
Similarly, given that the local minimum occurs when x = 5, substitute this value into the equation:
6(5)^2 + 2a(5) + b = 0
150 + 10a + b = 0 -- (equation 2)
We now have a system of two equations (equation 1 and equation 2) with two unknowns (a and b).
Solving the system of equations can help us find the values of a and b.
Using any preferred method of solving systems of linear equations, we can find that a = -6 and b = 6.
Therefore, the values of a and b for the function y = 2x^3 + ax^2 + bx + 36 are a = -6 and b = 6.
at both max and min, the derivative is zero, so
dy/dx = 6x^2 + 2ax + b
at x=-4,
96 - 8a + b = 0
at x=5
150 + 10a + b = 0
subtract them
54 + 16a = 0
a = - 27/8
sub that back into one of the above equations to find b