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.
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.
1 answer