Given that 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 maxima, minima, or neither.
f' = 3x^2+2ax+b
Now just find a and b so that f' has factors (x-1)(x-5)
Then you can determine the rest using f".
To find the values of a and b, we need to use the concept of critical points. Critical points occur when the derivative of a function is equal to zero or is undefined. In this case, f(x) has critical points at x = 1 and x = 5.
Step 1: Find the derivative of f(x) with respect to x.
f'(x) = 3x^2 + 2ax + b
Step 2: Set the derivative equal to zero and solve for x.
0 = 3x^2 + 2ax + b
Since we are given that x = 1 and x = 5 are critical points, we can set up two equations as follows:
Equation 1: 0 = 3(1)^2 + 2a(1) + b => 0 = 3 + 2a + b (substituting x=1)
Equation 2: 0 = 3(5)^2 + 2a(5) + b => 0 = 75 + 10a + b (substituting x=5)
Step 3: Solve the system of equations using the given critical points.
Using the equations obtained in Step 2, we can solve for a and b:
From Equation 1: 0 = 3 + 2a + b
Rearranging the equation: b = -3 - 2a
Substituting this value of b into Equation 2:
0 = 75 + 10a + (-3 - 2a)
Simplifying the equation:
0 = 72 + 8a
Solving for a:
-8a = 72
a = -9
Substituting the value of a into Equation 1:
b = -3 - 2(-9)
b = -3 + 18
b = 15
So, the values of a and b are a = -9 and b = 15.
Step 4: Classify the critical points as maxima, minima, or neither.
To classify the critical points, we need to determine the nature of these points. We can use the second derivative test.
Step 4.1: Find the second derivative of f(x).
f''(x) = 6x + 2a
Step 4.2: Evaluate the second derivative at x = 1 and x = 5.
For x = 1:
f''(1) = 6(1) + 2(-9) = 6 - 18 = -12
For x = 5:
f''(5) = 6(5) + 2(-9) = 30 - 18 = 12
Step 4.3: Analyze the second derivative test.
If f''(1) > 0, then the critical point at x = 1 is a local minimum.
If f''(1) < 0, then the critical point at x = 1 is a local maximum.
If f''(5) > 0, then the critical point at x = 5 is a local minimum.
If f''(5) < 0, then the critical point at x = 5 is a local maximum.
Based on the results:
- Since f''(1) = -12 < 0, the critical point at x = 1 is a local maximum.
- Since f''(5) = 12 > 0, the critical point at x = 5 is a local minimum.
Therefore, the critical point at x = 1 is a local maximum, and the critical point at x = 5 is a local minimum.