SOLUTION: Determine whether the Mean Value theorem can be applied to f on the closed interval f(x) = sqrt2-x, [−7, 2] If the Mean Value Theorem can be applied, find all values of

Algebra ->  Trigonometry-basics -> SOLUTION: Determine whether the Mean Value theorem can be applied to f on the closed interval f(x) = sqrt2-x, [−7, 2] If the Mean Value Theorem can be applied, find all values of      Log On


   

Question 1187614: Determine whether the Mean Value theorem can be applied to f on the closed interval
f(x) = sqrt2-x, [−7, 2]
If the Mean Value Theorem can be applied, find all values of c in the open interval (a, b)such that f '(c) = f(b) − f(a)/b − a. (Enter your answers as a comma-separated list. If the Mean Value Theorem cannot be applied, enter NA.)
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
You determine whether it satisfies the hypotheses by determining whether
f%28x%29=sqrt%282-x%29+is continuous on the interval [-7,2] and differentiable on the interval (-7,2). (Those are the hypotheses of the Mean Value Theorem)
You find the c mentioned in the conclusion of the theorem by solving
f'%28x%29=%28f%282%29-f%28-7%29%29%2F%282-%28-7%29%29+
f'%28x%29=%28f%282%29-f%28-7%29%29%2F%282%2B7%29+ on the interval (-7,2).
f is continuous on its domain, which includes [-7,2]
%28d%2Fdx%29%28sqrt%282+-+x%29%29+=+-1%2F%282sqrt%282+-+x%29%29
f'%28x%29 exists for all x%3C2 so it exists for all x in (-7,2)
Therefore this function satisfies the hypotheses of the Mean Value Theorem on this interval.
To find c+solve the equation
f'%28x%29=%28f%282%29-f%28-7%29%29%2F%282%2B7%29 ... f%282%29=sqrt%282-2%29+=0, f%28x%29=sqrt%282%2B7%29=sqrt%289%29=3+
f'%28c%29=-1%2F%282sqrt%282+-+c%29%29
discard any solutions outside (-7,2), and you get
-1%2F%282sqrt%282+-c%29%29=%280-3%29%2F%282%2B7%29+
3%2Fsqrt%282+-+c%29+=+2
3%2F2=sqrt%282+-+c%29
9%2F4=2+-+c
c=2+-+9%2F4
c=-1%2F4