A dunction is continious on the closed interval [-3,3] such that of f(-3)=4 and f(3)=1. The functions F' anf F'' have the properties shown below
Ok there's supposed to be a table but you cant really see it
-3<x<-1
F'(x) is positive
F''(X) is positive
X=-1
f'(x)and f''(x) failed to exist
-1<x<1
f'(x) is negative
f''(x) is positive
x=0
F'(x) and f''(x) are 0
1<x<3 f'(x) and f''(x) are negative
a) what are the x-coordinates of all the absolute maxes and minimu points of f on the inerval [-3,3] Justify your answer
b)what are the x-coordinateds of all points of inflection of f on the interval [-3,3]? Justify your answer
To find the x-coordinates of the absolute maximum and minimum points of f on the interval [-3,3], we need to identify the critical points and endpoints on that interval.
Step 1: Identify the critical points.
Critical points occur where f'(x) = 0 or f'(x) does not exist.
From the given information:
- At x = -3, f'(-3) = positive (as stated in the table), so it is not a critical point.
- At x = -1, f'(-1) does not exist, so it is a critical point.
- At x = 0, f'(0) = 0, so it is a critical point.
- At x = 1, f'(1) is negative (as stated in the table), so it is a critical point.
- At x = 3, f'(3) = negative (as stated in the table), so it is a critical point.
Step 2: Identify the endpoints.
The endpoints of the interval [-3,3] are -3 and 3.
Step 3: Evaluate f(x) at the critical points and endpoints.
Evaluate f(x) at x = -3, -1, 0, 1, and 3.
- At x = -3, f(-3) = 4 (given).
- At x = -1, find f(-1).
- At x = 0, find f(0).
- At x = 1, find f(1).
- At x = 3, f(3) = 1 (given).
Now, we need to analyze the values of f(x) at these points to determine the absolute maximum and minimum.
a) Absolute Maximum and Minimum Points:
- Compare the values of f(x) at critical points and endpoints.
- The largest value among these is the absolute maximum, and the smallest value is the absolute minimum.
Let us continue evaluating f(x) at x = -1, 0, and 1.
At x = -1, f'(-1) does not exist, so we cannot directly determine if it is the maximum or minimum. We need to analyze further.
From the table, when -1 < x < 1, f'(x) is negative, which means f(x) is decreasing in that interval. Hence, f(-1) should be greater than f(-3) = 4, making it a maximum point.
At x = 0, f'(0) = 0. This means that x = 0 is a critical point but not an extremum.
At x = 1, f'(1) is negative. This indicates that x = 1 is a minimum point.
Finally, compare the values at the endpoints.
f(-3) = 4, f(3) = 1.
From the above analysis, we conclude:
- The absolute maximum point of f on the interval [-3, 3] is at x = -1.
- The absolute minimum point of f on the interval [-3, 3] is at x = 1.
b) Points of Inflection:
To find the points of inflection, we need to identify the values of x where f''(x) changes its sign or f''(x) equals zero.
From the given information:
- At x = -3, no statement is given about f''(x), so we cannot determine if it is a point of inflection.
- At x = -1, f''(-1) does not exist, so it is not a point of inflection.
- At x = 0, f''(0) is given as zero, making x = 0 a possible point of inflection.
- At x = 1, f''(1) is given as positive, indicating that x = 1 is not a point of inflection.
- At x = 3, no statement is given about f''(x), so we cannot determine if it is a point of inflection.
Therefore, the x-coordinate of the potential point of inflection is x = 0.
In summary:
a) The x-coordinate of the absolute maximum point of f on the interval [-3, 3] is -1, and the x-coordinate of the absolute minimum point is 1.
b) The x-coordinate of the potential point of inflection of f on the interval [-3, 3] is 0.