Let f(x) = (x- 3)^-2 Find all values of c in (1, 4) such that f(4) - f(1) = f?(c
ID: 2867075 • Letter: L
Question
Let f(x) = (x- 3)^-2 Find all values of c in (1, 4) such that f(4) - f(1) = f?(c)(4 - 1). (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.) Based off of this information, what conclusions can be made about the Mean Value Theorem? This contradicts the Mean Value Theorem since f satisfies the hypotheses on the given interval but there does not exist any con (1, 4) such that f?(c) =f(4-1)/4-1 This does not contradict the Mean Value Theorem since f is not continuous at x = 3. This does not contradict the Mean Value Theorem since f is continuous on (1, 4), and there exists a c on (1, 4) such that f?(c) =f(4-1)/4-1 This contradicts the Mean Value Theorem since there exists a c on (1, 4) such that f?(c) =f(4-1)/4-1, but f is not continuous at x = 3. Nothing can be concluded.Explanation / Answer
(1,4) is an open interval -- that the values 1 and 4 are not included in the interval.
f(x) = 1/(x - 3)^2
has an asymptote at x = 3, so the function is undefined and discontinuous at that point. The derivative is
f '(x) = -2(x - 3)/(x - 3)^4 = -2/(x - 3)^3
It also is undefined and discontinuous when x = 3. So the function and its derivative have a discontinuity in the interval.
f(4) = 1 and f(1) = 0.25, so [f(4) - f(1)]/(4 - 1) = 0.25
Let's see where the derivative is equal to 0.25.
-2/(x - 3)^3 = 0.25
(x - 3)^3 = -2/0.25 = -8
Take the cube root of both sides,
x - 3 = -2
x = 1
So the derivative is 0.25 when x = 1, but, since we have an open interval, 1 is not in the interval.
This does not contradict the mean value theorem since f is not continuous at x=3
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