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Anonymous
Anonymous asked in Science & MathematicsMathematics · 10 years ago

# help!calculus!concavity!?

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 (__ , __)

thx all!!!

### 4 Answers

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• Favorite Answer

Not sure exactly, but concave is like this: (__)

and convex is like this: )__(

Hope that helps.

• 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) âª (5/4,â) concave up

(0,5/4) is concave down

inflection points at x = 0 and x = 5/4

• 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) âª(5/4,+â) and down on (0, 5/4)

Source(s): Accounting major
• Anonymous
10 years ago

http://en.wikipedia.org/wiki/Inflection_point

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.

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