given tat F(x)=x^3+ax^2+bx has critical points at x=1 and x=5 find a and b and classify the critical points as max, min, or neither.
To find the values of a and b, we can use the information provided about the critical points.
When F(x) has a critical point at x=1, it means that the derivative of F(x) is equal to zero at x=1:
F'(1) = 0
To find F'(x), we can use the power rule for differentiation:
F'(x) = 3x^2 + 2ax + b
Substituting x=1 into the derivative equation, we get:
3(1)^2 + 2a(1) + b = 0
Simplifying:
3 + 2a + b = 0 (Equation 1)
Similarly, when F(x) has a critical point at x=5, the derivative of F(x) is equal to zero:
F'(5) = 0
Substituting x=5 into the derivative equation:
3(5)^2 + 2a(5) + b = 0
Simplifying:
75 + 10a + b = 0 (Equation 2)
Now, we have a system of two equations (Equation 1 and Equation 2) with two unknowns (a and b). We can solve this system to find the values of a and b.
By subtracting Equation 1 from Equation 2, we eliminate b:
(75 + 10a + b) - (3 + 2a + b) = 0
75 + 10a + b - 3 - 2a - b = 0
7 + 8a = 0
8a = -7
a = -7/8
Substituting this value of a back into Equation 1, we can solve for b:
3 + 2(-7/8) + b = 0
3 - 7/4 + b = 0
12/4 - 7/4 + b = 0
5/4 + b = 0
b = -5/4
Therefore, the values of a and b are a = -7/8 and b = -5/4.
To classify the critical points, we can use the second derivative test. The second derivative of F(x) is:
F''(x) = 6x + 2a
Substituting x=1 and x=5 into the second derivative:
F''(1) = 6(1) + 2(-7/8) = 6 - 7/4 = 17/4 > 0
F''(5) = 6(5) + 2(-7/8) = 30 - 7/4 = 113/4 > 0
Since the second derivative is positive at both critical points, the critical points are classified as minima.