The derivative of a function is f'(x)=x(x+2)(x-5). Find the value of x at each point where f has a
(a) local maximum,
(b) local minimum, or
(c) point of inflection
there are max/min at x = -2, 0, 5
f''(x) = 3x^2 - 6x - 10
at -2, 0, -5, f'' = 14, -10, 95
so, min at x = -2
max at x = 0
min at x = 5
inflection at 1±√39/3
To find the local maximum, local minimum, and points of inflection of a function, we need to determine the critical points and use the second derivative test. Here's how you can do it step by step:
Step 1: Find the critical points by setting the derivative equal to zero and solving for x.
Given that f'(x) = x(x+2)(x-5), to find the critical points, we need to solve the equation f'(x) = 0.
x(x+2)(x-5) = 0
We can see that the product of three factors is equal to zero. This means that at least one of the factors must be equal to zero. So we have three possible cases:
Case 1: x = 0
Case 2: x + 2 = 0 => x = -2
Case 3: x - 5 = 0 => x = 5
These values of x (0, -2, and 5) are the critical points of the function.
Step 2: Classify the critical points using the second derivative test.
To determine whether each critical point is a local maximum, local minimum, or point of inflection, we need to consider the concavity of the function using the second derivative test.
The concavity of the function can be determined by evaluating the second derivative of the function, f''(x).
To find the second derivative, we need to differentiate f'(x) = x(x+2)(x-5) with respect to x.
f''(x) = (x+2)(x-5) + x(x-5) + x(x+2)
Expanding and simplifying the above expression, we get:
f''(x) = x^2 - 5x + 2x - 10 + x^2 - 5x + x^2 + 2x
f''(x) = 3x^2 - 6x - 10
Step 3: Evaluate f''(x) at the critical points.
Now, we will substitute the critical points (0, -2, and 5) into the second derivative equation, f''(x), to determine the concavity at each point.
For x = 0:
f''(0) = 3(0)^2 - 6(0) - 10 = -10
For x = -2:
f''(-2) = 3(-2)^2 - 6(-2) - 10 = 10
For x = 5:
f''(5) = 3(5)^2 - 6(5) - 10 = 55
Step 4: Analyze the concavity and classify the critical points.
To classify each critical point, we analyze the concavity at that point.
For a local maximum, the concavity changes from positive to negative.
For a local minimum, the concavity changes from negative to positive.
For a point of inflection, the concavity does not change.
Analyzing the results:
- At x = 0: f''(0) < 0, indicating a local maximum.
- At x = -2: f''(-2) > 0, indicating a local minimum.
- At x = 5: f''(5) > 0, indicating a local minimum.
Therefore, the function f(x) has a local maximum at x = 0 and local minimums at x = -2 and x = 5.