Asked by Shalma
                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 :)
 
            
            
        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 :)
Answers
                    Answered by
            Van
            
    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).
    
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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