Use a sign chart to determine the intervals on which f(x)=4x^3-x^4 is concave up and concave up and concave down, and identify the locations of any inflection points.
To determine the intervals on which the function f(x) = 4x^3 - x^4 is concave up or concave down, we need to analyze its second derivative. The concavity of a function can change at the points where the second derivative is equal to zero or undefined, known as inflection points.
Let's first find the first and second derivatives of f(x).
First derivative:
f'(x) = 12x^2 - 4x^3
Second derivative:
f''(x) = 24x - 12x^2
To create a sign chart for f''(x), we need to determine the critical points where f''(x) = 0 or f''(x) is undefined.
Setting f''(x) = 0:
24x - 12x^2 = 0
12x(2 - x) = 0
From this equation, we find two critical points: x = 0 and x = 2.
Now, we need to test intervals to the left and right of these critical points to determine the concavity. We can choose any number within each interval and plug it into the second derivative.
For x < 0: (Pick x = -1)
f''(-1) = 24(-1) - 12(-1)^2 = -24 + 12 = -12 (negative)
For 0 < x < 2: (Pick x = 1)
f''(1) = 24(1) - 12(1)^2 = 24 - 12 = 12 (positive)
For x > 2: (Pick x = 3)
f''(3) = 24(3) - 12(3)^2 = 72 - 108 = -36 (negative)
Now let's summarize our findings:
Interval (-∞, 0): f''(x) < 0, thus f(x) is concave down.
Interval (0, 2): f''(x) > 0, thus f(x) is concave up.
Interval (2, ∞): f''(x) < 0, thus f(x) is concave down.
Therefore, f(x) is concave up on the interval (0, 2) and concave down on the intervals (-∞, 0) and (2, ∞).
Next, to identify the locations of inflection points, we look for the x-values where the concavity changes, which is when f''(x) = 0 or f''(x) is undefined.
In this case, we found two critical points: x = 0 and x = 2. So these are potential inflection points.
To determine if they are indeed inflection points, we examine the behavior of the function on both sides of each critical point.
For x < 0: (Pick x = -1)
f(-1) = 4(-1)^3 - (-1)^4 = -4 + 1 = -3 (negative)
For 0 < x < 2: (Pick x = 1)
f(1) = 4(1)^3 - (1)^4 = 4 - 1 = 3 (positive)
For x > 2: (Pick x = 3)
f(3) = 4(3)^3 - (3)^4 = 108 - 81 = 27 (positive)
Based on the behavior of f(x) on either side of the potential inflection points, we can conclude that x = 0 is not an inflection point since the function changes concavity without crossing the x-axis. However, x = 2 is an inflection point as the function changes concavity and crosses the x-axis.
To summarize:
- f(x) is concave up on the interval (0, 2) and concave down on the intervals (-∞, 0) and (2, ∞).
- The inflection point occurs at x = 2.
By creating a sign chart and analyzing the behavior of the function and its derivatives at critical points, we were able to determine the intervals of concavity and detect the inflection point.