Verify that the hypotheses of the Mean-Value Theorem are satisfied for
f(x) = √(16-x^2 ) on the interval [-4,1]
and find all values of C in this interval that satisfy the conclusion of the theorem.
Briefly, the MVT states that given an arc of a differentiable curve, there is at least one point on that arc at which the slope of the curve is equal to the "average" derivative of the arc, namely the chord between the end-points.
First, see if the function is differentiable between the end-points:
f(x)=√(16-x²) on [-4,1]
f(x) exists and is continuous at all points between -4 and 1. There are no vertical asymptotes or other discontinuities. There is a maximum at x=0.
The slope of the chord
= (f(1)-f(-4))/(1-(-4))
= (√15 - √0)/5
= √15 /5
Differentiate f(x) to get f'(x)
f'(x)= -x/&radic(16-x²)
If the mean-value theorem is valid for this interval, at least one solution exists for f'(x)=√15 / 5
Let's see:
-x/&radic(16-x²) = √15 /5
cross multiply:
-5x = √(15*16-15x²)
Square both sides (and check all roots afterwards):
25x² = 240-15x²
x²=6
x=±√6
since x=+√6 > 1 and is outside of the given interval, it is rejected.
So x=-√6
I will leave it to you to check that the mean-value theorem is satisfied.
To verify that the hypotheses of the Mean Value Theorem (MVT) are satisfied for a given function on a specific interval, we need to check two conditions:
1. Continuity: The function f(x) must be continuous on the interval [a, b].
2. Differentiability: The function f(x) must be differentiable on the open interval (a, b).
Let's analyze these conditions for the given function f(x) = √(16-x^2) on the interval [-4, 1].
1. Continuity:
To ensure continuity, we need to check if the function is defined and continuous at both endpoints of the interval.
At x = -4:
f(-4) = √(16 - (-4)^2) = √(16 - 16) = √0 = 0
At x = 1:
f(1) = √(16 - 1^2) = √(16 - 1) = √15
Since f(x) is defined and continuous at both endpoints, the first condition of continuity is satisfied.
2. Differentiability:
To check differentiability, we need to verify if the derivative exists and is finite on the open interval (-4, 1).
The derivative of f(x) = √(16-x^2) can be found by applying the Chain Rule:
f'(x) = -x / √(16 - x^2)
There is no issue with the denominator being zero within the interval (-4, 1). Hence, the derivative exists and is finite on the open interval (-4, 1). Therefore, the second condition of differentiability is satisfied.
Since both the conditions of continuity and differentiability are satisfied, we can conclude that the hypotheses of the Mean Value Theorem are satisfied for the function f(x) = √(16-x^2) on the interval [-4, 1].
To find all values of c that satisfy the conclusion of the theorem, we can use the intermediate value theorem, which states that for any continuous function on a closed interval [a, b], there exists at least one c in the open interval (a, b) such that f'(c) = (f(b) - f(a)) / (b - a).
The conclusion of the Mean Value Theorem states that there exists at least one c in the interval (-4, 1) such that f'(c) = (f(1) - f(-4)) / (1 - (-4)).
Substituting the values:
f(1) = √15
f(-4) = 0
f'(c) = (f(1) - f(-4)) / (1 - (-4))
f'(c) = (√15 - 0) / (1 + 4)
f'(c) = √15 / 5
f'(c) = (1/5) √15
Therefore, the value of c that satisfies the conclusion of the theorem is any value in the interval (-4, 1) for which f'(c) = (1/5) √15.