If f(x) is differentiable for the closed interval [-3, 2] such that f(-3) = 4 and f(2) = 4, then there exists a value c, -3 < c < 2 such that (4 points)
If f(x) = ι(x2 - 8)ι, how many numbers in the interval 0 ≤ x ≤ 2.5 satisfy the conclusion of the mean value theorem? (
To find out how many numbers in the interval 0 ≤ x ≤ 2.5 satisfy the conclusion of the mean value theorem, we need to determine if there exists a value c, -3 < c < 2, such that f'(c) = (f(2) - f(-3)) / (2 - (-3)).
First, let's find f(x) by plugging in the given function f(x) = ι(x^2 - 8)ι into the interval [-3, 2]:
f(-3) = ι((-3)^2 - 8)ι = ι(9 - 8)ι = ι(1)ι = 1
f(2) = ι(2^2 - 8)ι = ι(4 - 8)ι = ι(-4)ι = 4
Next, let's find f'(x) by differentiating f(x) with respect to x. Since f(x) = ι(x^2 - 8)ι makes a sharp turn at x = 0, we need to use the chain rule when differentiating.
f'(x) = ι(x^2 - 8)ι'
= 2x(i(x^2 - 8))'
= 2xi'(x^2 - 8) + 2x(i(x^2 - 8))'
= 2xi'(x^2 - 8) + 2x(1)
= 2x(i'(x^2 - 8) + 1)
Now, let's find the critical points of f(x) by setting f'(x) = 0:
2x(i'(x^2 - 8) + 1) = 0
Since 2x is never equal to zero in the interval [-3, 2], we can ignore it.
i'(x^2 - 8) + 1 = 0
i'(x^2 - 8) = -1
Since i(x) is the absolute value function, its derivative is either -1 or 1, depending on whether x^2 - 8 is negative or positive.
If x^2 - 8 < 0, then i'(x^2 - 8) = -1
If x^2 - 8 > 0, then i'(x^2 - 8) = 1
To determine whether x^2 - 8 is negative or positive in the interval [-3, 2], we can examine its sign at the critical points -3 and 2.
For x = -3:
(-3)^2 - 8 = 1 - 8 = -7
Since x^2 - 8 is negative for x = -3, i'(x^2 - 8) = -1.
For x = 2:
2^2 - 8 = 4 - 8 = -4
Since x^2 - 8 is negative for x = 2, i'(x^2 - 8) = -1.
Therefore, we can conclude that i'(x^2 - 8) = -1 for the entire interval [-3, 2].
Now, substituting this value into the equation i'(x^2 - 8) = -1:
-i'(x^2 - 8) = -(-1)
i'(x^2 - 8) = 1
Since we want to find a value c, -3 < c < 2, such that f'(c) = (f(2) - f(-3)) / (2 - (-3)):
f'(c) = 1
Finally, we need to count the number of numbers in the interval 0 ≤ x ≤ 2.5 that satisfy f'(c) = 1. To do this, we need to determine if f'(x) = 1 for any x in the interval.
Since f'(x) = i'(x^2 - 8) = 1, we need to solve the equation:
1 = 1
This equation is always satisfied, regardless of the value of x.
Therefore, for any value of x in the interval 0 ≤ x ≤ 2.5, f'(c) = 1, and thus, the conclusion of the mean value theorem is satisfied for all numbers in the given interval.
Hence, all numbers in the interval 0 ≤ x ≤ 2.5 satisfy the conclusion of the mean value theorem.
To apply the Mean Value Theorem on the interval [-3, 2], we need f(x) to be a continuous function on the closed interval [-3, 2] and differentiable on the open interval (-3, 2).
Given that f(x) = ι(x^2 - 8)ι, we have:
-3 ≤ x ≤ 2 -- (1)
Now, let's check if f(x) is continuous on the closed interval [-3, 2].
For a function to be continuous, f(x) must be defined and the limit of f(x) as x approaches any point within the interval must exist and be equal to the value of the function at that point.
In our case, f(x) = ι(x^2 - 8)ι is defined for all real numbers. So, f(x) is defined for all x in [-3, 2].
Now, let's check if the limit of f(x) as x approaches any point within the interval exists and is equal to the value of the function at that point.
For x in (-∞, -3) ∪ (-3, ∞), we have f(x) = x^2 - 8 since the absolute value function ιxι becomes x when x is negative.
So, the limit of f(x) as x approaches any point within the interval (-∞, -3) ∪ (-3, ∞) is the same as the limit of x^2 - 8, which exists and is equal to the value of f(x) at those points.
Similarly, for x in [-3, 2], the absolute value function ιx^2 - 8ι becomes -(x^2 - 8).
So, the limit of f(x) as x approaches any point within the interval [-3, 2] is the same as the limit of -(x^2 - 8), which exists and is equal to the value of f(x) at those points.
Therefore, f(x) is continuous on the closed interval [-3, 2].
Next, let's check if f(x) is differentiable on the open interval (-3, 2).
To be differentiable at any point c within an interval, f(x) must be differentiable at each individual point c within the interval.
Since f(x) = x^2 - 8 for x in (-∞, -3) ∪ (-3, ∞), f(x) is differentiable at these points.
For x in [-3, 2], f(x) = -(x^2 - 8), and (x^2 - 8) is differentiable for all x in [-3, 2]. Therefore, f(x) is differentiable at these points as well.
Therefore, f(x) is differentiable on the open interval (-3, 2).
Now, to apply the Mean Value Theorem, we need to verify if f(x) satisfies the conditions of the Mean Value Theorem.
1. f(x) is continuous on the closed interval [-3, 2]: Checked and satisfied.
2. f(x) is differentiable on the open interval (-3, 2): Checked and satisfied.
3. f(-3) = 4: Given information, which is satisfied.
4. f(2) = 4: Given information, which is satisfied.
All the conditions of the Mean Value Theorem are satisfied.
According to the Mean Value Theorem, there exists a value c, -3 < c < 2, such that the instantaneous rate of change of f(x) at c is equal to the average rate of change of f(x) over the interval [-3, 2].
To find the number of values c, -3 < c < 2, that satisfy the conclusion of the Mean Value Theorem, we need to determine if f(x) has any points where the instantaneous rate of change is equal to the average rate of change over the interval [-3, 2].
The average rate of change of f(x) over the interval [-3, 2] is given by:
Average rate of change = (f(2) - f(-3))/(2 - (-3))
= (4 - 4)/(2 - (-3))
= 0/5
= 0
We need to find the values of c where the instantaneous rate of change of f(x) is equal to 0.
Taking the derivative of f(x) = ι(x^2 - 8)ι, we get:
f'(x) = 2x for x > 0
-2x for x < 0
Since we are interested in x values where -3 < x < 2, we consider the case where x > 0. In this case, f'(x) = 2x.
Setting f'(x) = 0, we have:
2x = 0
x = 0
Therefore, the only value of c, -3 < c < 2, that satisfies the conclusion of the Mean Value Theorem is c = 0.
In summary, there exists one value c, -3 < c < 2, such that the conclusion of the Mean Value Theorem is satisfied for the function f(x) = ι(x^2 - 8)ι.
study any proof of the MVT or Rolle's Theorem for the first one.
For the 2nd one, since f satisfies the conditions of the MVT, and is concave down over the whole interval, there is but one number c in the interval.