Test f(x)=2x^4-5x^3+5 for concavity and inflection points.
a) f has inflection points when x=____and x=_____(list the smaller value first)
b) f is concave up on ( -∞,__) ∪ (___,+∞) and its concave down on (__ , __)
- JJLv 710 years agoFavorite Answer
Not sure exactly, but concave is like this: (__)
and convex is like this: )__(
Hope that helps.
- 10 years ago
f(x) = 2x⁴ - 5xÂ³ + 5
f'(x) = 8xÂ³ - 15xÂ²
f"(x) = 24xÂ² - 30x
(using 2nd derivative to determine concavity and inflection points)
set the second derivative to zero and solve for x, the inflection points
24xÂ² - 30x = 0
6x(4x-5) = 0
6x = 0
x = 0
4x-5 = 0
4x = 5
x = 5/4
picking any point in between -â and 0, 0 and 5/4, 5/4 and â to test concavity.
if it's negative it's concave down and if it's costive it's concave up.
in this case
-â < -1 < 0 â f"(-1) > 0 so (-â,0) is concave up
0 < 1 <5/4 â f"(1) < 0 so (0,5/4) is concave down
5/4 < 2 < â â f"(2) > 0 so (5/4,â) is concave up
(-â,0) âª (5/4,â) concave up
(0,5/4) is concave down
inflection points at x = 0 and x = 5/4
- 10 years ago
Find the double prime of the function, which is f ''(x) = 24x^2-30x. Set that to zero to get your inflection points and concavity.
a) (0,5) and (5/4, 15/128)
b) up on (-â,0) âª(5/4,+â) and down on (0, 5/4)Source(s): Accounting major
- Anonymous10 years ago
If x is an inflection point for f then the second derivative, f″(x), is equal to zero if it exists, but this condition does not provide a sufficient definition of a point of inflection
Read that, have a go. Take values either side to check if its going up or down.
Lastly, plot a few values to see what the function looks like on paper. Good luck.