Let f (x) = x3 ��3x2 +5x+2. Find the open intervals on which f is concave up (down). Then determine the
x-coordinates of all inflection points of f .
1. f is concave up on the intervals
2. f is concave down on the intervals
3. The inflection points occur at x =
To find the open intervals on which f is concave up or down, we need to determine the second derivative of the function f(x) and analyze its sign changes.
Step 1: Find the first derivative of f(x)
Differentiate f(x) with respect to x:
f'(x) = 3x^2 - 6x + 5
Step 2: Find the second derivative of f(x)
Differentiate f'(x) with respect to x:
f''(x) = 6x - 6
Step 3: Set f''(x) equal to 0 and solve for x
To find the intervals where the concavity changes, we need to find the x-values where the second derivative is equal to zero:
6x - 6 = 0
6x = 6
x = 1
Step 4: Analyze the sign changes
Now we can determine the intervals on which f is concave up or down by analyzing the sign changes of the second derivative (f''(x)).
For x < 1:
Choose a test value less than 1, let's say x = 0.
Substitute x = 0 into f''(x):
f''(0) = 6(0) - 6
f''(0) = -6
Since f''(0) is negative, this means f is concave down on the interval x < 1.
For x > 1:
Choose a test value greater than 1, let's say x = 2.
Substitute x = 2 into f''(x):
f''(2) = 6(2) - 6
f''(2) = 6
Since f''(2) is positive, this means f is concave up on the interval x > 1.
Step 5: Determine the inflection points
To find the x-coordinates of the inflection points, we need to find the x-values where the concavity changes. In this case, we have one inflection point at x = 1.
Therefore, to summarize:
1. f is concave up on the intervals: x > 1
2. f is concave down on the intervals: x < 1
3. The inflection point occurs at x = 1.