Find the critical numbers of the function and determine local maximum and minimum
values of f
(a) f(x) = 2x
3 − 3x
2 − 36x
(b) f(t) = |3t − 4|
To find the critical numbers of a function and determine its local maximum and minimum values, we need to follow these steps:
(a) f(x) = 2x^3 - 3x^2 - 36x
Step 1: Find the derivative of the function f(x).
The derivative of f(x) is found by taking the derivative of each term separately. Since the function is polynomial, we can apply the power rule.
f'(x) = 6x^2 - 6x - 36
Step 2: Set the derivative equal to zero and solve for x to find the critical numbers.
To find the critical numbers, we need to solve the equation f'(x) = 0.
6x^2 - 6x - 36 = 0
Factor out 6 from each term:
6(x^2 - x - 6) = 0
Now, solve the quadratic equation either by factoring, completing the square, or using the quadratic formula:
x^2 - x - 6 = 0
Factoring gives us:
(x - 3)(x + 2) = 0
So, x = 3 or x = -2.
Step 3: Determine the local maximum and minimum values.
To determine the local maximum and minimum values, we need to analyze the behavior of the function around the critical numbers.
We can use the second derivative test to determine whether each critical number corresponds to a local maximum or minimum.
The second derivative f''(x) is found by taking the derivative of f'(x):
f''(x) = 12x - 6
Evaluate f''(x) at each critical number:
f''(3) = 12(3) - 6 = 30
f''(-2) = 12(-2) - 6 = -30
If f''(x) is positive at a critical number, then it corresponds to a local minimum. If f''(x) is negative at a critical number, then it corresponds to a local maximum.
Since f''(3) is positive (30), x = 3 corresponds to a local minimum.
Since f''(-2) is negative (-30), x = -2 corresponds to a local maximum.
Therefore, the critical numbers of f(x) = 2x^3 - 3x^2 - 36x are x = 3 and x = -2. The local minimum occurs at x = 3, and the local maximum occurs at x = -2.
(b) f(t) = |3t - 4|
Step 1: Find the derivative of the function f(t).
Since the function involves an absolute value, we need to consider the derivative on different intervals where the expression inside the absolute value changes sign.
For 3t - 4 >= 0 (when 3t - 4 is nonnegative), the function f(t) can be written as:
f(t) = 3t - 4
Taking the derivative of f(t), we get:
f'(t) = 3
For 3t - 4 < 0 (when 3t - 4 is negative), the function f(t) can be written as:
f(t) = -(3t - 4) = -3t + 4
Taking the derivative of f(t), we get:
f'(t) = -3
Step 2: Determine the critical numbers.
Since the derivative is constant for both cases (3 and -3), there are no critical numbers for this function.
Step 3: Determine the local maximum and minimum values.
Since there are no critical numbers, there are no local maximum or minimum values for this function.
Therefore, the function f(t) = |3t - 4| has no critical numbers, and therefore, no local maximum or minimum values.