f(x)=[3x^13]e^(−7x)
On which intervals is the function concave up and concave down?
I know when you find the 2nd derivative, the x-values are 0, 1.34206, and 2.37222, and that the 2nd derivative > 0 at x=1.34205 and < 0 at 2.37222.
How does this tell me the intervals where it's concave up and down?
concave down where f" < 0
better review your text, where it is surely explained and illustrated.
f" < 0 means that the slope is decreasing. So, f is either rising either rising more slowly or falling more quickly. Consider a parabola which opens downward.
so would concave up be (1.34205, 2.37222)?
Judging from the graph at
http://www.wolframalpha.com/input/?i=[3x^13]e^%28%E2%88%927x%29+for+x%3D0..3
I'd say it is concave down in that interval. In fact, since
f"(x) = 3x^11 e^-7x (49x^2-182x+156)
if the roots are where you say they are, then since the quadratic factor is a parabola which opens up, it is negative between the roots, meaning that f(x) is indeed concave down in that interval.
it still says I'm wrong, but thanks for trying
To determine the intervals on which a function is concave up or concave down, you need to analyze the sign of the second derivative.
When the second derivative is greater than zero, f''(x) > 0, it indicates that the function is concave up. This means that the graph of the function is curving upward like a "U" shape.
When the second derivative is less than zero, f''(x) < 0, it indicates that the function is concave down. This means that the graph of the function is curving downward like an inverted "U" shape.
In your case, you found that the second derivative is positive at x = 1.34205 and negative at x = 2.37222.
To determine the intervals where the function is concave up or down, you can use the following steps:
1. Identify the critical points: These are the x-values where the second derivative changes sign. In your case, you mentioned x = 1.34205 and x = 2.37222.
2. Divide the x-axis into intervals: Use the critical points to divide the x-axis into separate intervals. For your case, you can determine three intervals: (-∞, 1.34205), (1.34205, 2.37222), and (2.37222, +∞).
3. Evaluate the second derivative within each interval: Pick a random value within each interval and evaluate the second derivative f''(x) at that point to determine its sign.
- For (-∞, 1.34205), you can choose x = 0 as a convenient point. Evaluate f''(0) to determine its sign.
- For (1.34205, 2.37222), you can choose x = 2 as a convenient point. Evaluate f''(2) to determine its sign.
- For (2.37222, +∞), you can choose x = 3 as a convenient point. Evaluate f''(3) to determine its sign.
4. Analyze the signs: Determine whether the second derivative is greater than zero or less than zero within each interval based on your evaluations. This will tell you whether each interval is concave up or concave down.
For example, if f''(0) > 0, it means the interval (-∞, 1.34205) is concave up. If f''(2) < 0, it means the interval (1.34205, 2.37222) is concave down. And if f''(3) > 0, it means the interval (2.37222, +∞) is concave up.
By following these steps, you can determine on which intervals the function is concave up (U-shape) and concave down (inverted U-shape) based on the sign of the second derivative.