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 :)
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).
To identify the points of inflection, we need to analyze the behavior of the second derivative of the function f over the given intervals.
Looking at the second derivative table:
For 0 < x < 3, f'' is negative (-).
For 3 < x < 5, f'' is negative (-).
For 5 < x < 6, f'' is positive (+).
From this, we can conclude that f'' changes sign at x = 3 and x = 5.
Now, let's examine the behavior of the first derivative, f':
For 0 < x < 3, f' is positive (+).
For 3 < x < 5, f' is negative (-).
For 5 < x < 6, f' is negative (-).
From this, we can conclude that f' changes sign at x = 3 and x = 5.
Finally, let's analyze the behavior of the function f itself:
For 0 < x < 3, f is increasing.
For 3 < x < 5, f is decreasing.
For 5 < x < 6, f is decreasing.
From this, we can conclude that f changes concavity at x = 3 and x = 5.
So, we have identified the potential points of inflection at x = 3 and x = 5.
Now, we need to check if these points fulfill the condition for being points of inflection. A point of inflection occurs where f'' changes sign from positive to negative or from negative to positive.
At x = 3, f'' changes from negative to negative. This does not fulfill the condition for a point of inflection.
At x = 5, f'' changes from negative to positive. This fulfills the condition for a point of inflection.
Therefore, the only point of inflection is at (5, -1).
So, the answer is b. (5, -1) only.
To identify the points of inflection, we need to determine where the concavity of the graph changes. A point of inflection occurs when the concavity changes from concave up to concave down or vice versa.
From the given table, we see that the sign of the second derivative changes at x=3 and x=5. At x=3, the sign changes from negative to positive, indicating a change from concave down to concave up. At x=5, the sign changes from positive to negative, indicating a change from concave up to concave down.
Since the function is continuous on the interval [0,6], the points of inflection occur at x=3 and x=5, where the concavity changes.
So, the correct answer is:
c. (3,4) and (6, 3)
To recap the steps to finding the points of inflection:
1. Compute the second derivative of the function.
2. Determine the sign of the second derivative in each interval.
3. Identify the points where the sign of the second derivative changes.
4. These points represent the points of inflection.