Let f(x)=−x^4−8^x3+5^x+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 =
Did you reall mean 5^x, or 5x^2? I will assume the latter. Since
f" = -12x^2-48x+10
and that has two roots, those are where the inflection points lie.
To find the open intervals on which f is concave up or down and to determine the x-coordinates of all inflection points, we need to find the second derivative of f and analyze its sign changes.
1. To find the second derivative, we need to differentiate f(x) twice:
f(x) = -x^4 - 8x^3 + 5x + 2
f'(x) = -4x^3 - 24x^2 + 5
f''(x) = -12x^2 - 48x
2. To determine when f is concave up or down, we need to analyze the sign changes of the second derivative.
a. Set f''(x) = 0 and solve for x to find the potential inflection points:
-12x^2 - 48x = 0
-12x(x + 4) = 0
x = 0 or x = -4
b. Now we can construct a sign chart for f''(x) to analyze the intervals of concavity:
| -∞ | -4 | 0 | +∞ |
| - | + | - | + |
- For x < -4, f''(x) is positive (+), so f is concave up on this interval.
- For -4 < x < 0, f''(x) is negative (-), so f is concave down on this interval.
- For x > 0, f''(x) is positive (+) again, so f is concave up on this interval.
3. The inflection points occur at x = -4 and x = 0 since these are the x-coordinates where the concavity changes.
Therefore, the answers are:
1. f is concave up on the intervals (-∞, -4) and (0, +∞).
2. f is concave down on the interval (-4, 0).
3. The inflection points occur at x = -4 and x = 0.