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.
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?
4 answers
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.
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
0 answers