Posted by **Shalma** on Monday, January 3, 2011 at 9:23pm.

Suppose f is continuous on [0, 6] and satisfies the following

x 0 3 5 6

f -1 4 -1 -3

f' 5 0 -8 0

f" -1 -3 DNE 3

x 0<x<3 3<x<5 5<x<6

f' + - -

f" - - +

(they are both tables)

Identify the points of inflection

a. There are no points of inflection

b. (5, -1) only

c. (3,4) and (6, 3)

d. (3,4) only

e. (3,4),(6,-3), and (5, -1)

Please explain in details how you got the answer

THANK YOU :)

- Calculus -
**Van**, Sunday, January 9, 2011 at 8:00pm
Answer is b. (5,-1) only

Explanation:

Points of inflection are present when f"=0 or is undefined (DNE). From the first table, you can see that f" is not 0 for any values of x, but is undefined for x=5, so there can be a possible point of inflection at x=5. Then, look at the second table. Points of inflection are where concavity changes, where f" changes from one sign to another (- to +, or + to -). At x=5, the only possible point of inflection taken from the first table, f" does change from - to +, so there is a point of inflection at x=5. At x=5, f=-1, so the point is (5,-1) (aka the answer is b).

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