let f(x)= (x-3)^-2 Show that there is no value of c in (1,4) such that f(4)-f(1)= (f prime of c)(4-1). Why doesn't this contradict the mean value theorem.
To show that there is no value of c in (1, 4) such that f(4) - f(1) = f'(c)(4 - 1), we'll first find the expression for f'(x) and then evaluate f(4) - f(1).
Step 1: Find the derivative of f(x):
f(x) = (x - 3)^-2
Using the chain rule, we differentiate f(x) with respect to x:
f'(x) = -2(x - 3)^-3 * (1)
Simplifying this expression, we get:
f'(x) = -2/(x - 3)^3
Step 2: Evaluate f(4) and f(1):
f(4) = (4 - 3)^-2 = 1^(-2) = 1
f(1) = (1 - 3)^-2 = (-2)^-2 = 1/4
Step 3: Calculate f(4) - f(1):
f(4) - f(1) = 1 - 1/4 = 3/4
Step 4: Replace f'(c) and evaluate f'(c)(4 - 1):
Using the expression we found for f'(x), we replace x with c:
f'(c) = -2/(c - 3)^3
Now, calculate f'(c)(4 - 1):
f'(c)(4 - 1) = -2/(c - 3)^3 * 3 = -6/(c - 3)^3
Step 5: Compare f(4) - f(1) with f'(c)(4 - 1):
We need to show that f(4) - f(1) ≠ f'(c)(4 - 1) for any value of c in (1, 4).
From Step 3, we have f(4) - f(1) = 3/4.
From Step 4, we have f'(c)(4 - 1) = -6/(c - 3)^3.
Since 3/4 cannot equal -6/(c - 3)^3 for any value of c, we can conclude that there is no value of c in (1, 4) that satisfies f(4) - f(1) = f'(c)(4 - 1).
Now, as for why this doesn't contradict the Mean Value Theorem (MVT), it's because the function f(x) = (x - 3)^-2 is not continuous on the closed interval [1, 4]. The MVT can only be applied to continuous functions, and this function has a vertical asymptote at x = 3, which means it fails to meet the conditions required by the MVT.
To show that there is no value of c in (1, 4) such that f(4) - f(1) = f'(c)(4 - 1), we can start by calculating the values involved.
Given f(x) = (x - 3)^-2, we can find f'(x) by applying the power rule and the chain rule:
f'(x) = -2(x - 3)^-3 * (d/dx)(x - 3)
= -2(x - 3)^-3
Next, let's evaluate f(4) and f(1):
f(4) = (4 - 3)^-2
= 1
f(1) = (1 - 3)^-2
= 1/4
Plugging these values into the equation f(4) - f(1) = f'(c)(4 - 1), we have:
1 - 1/4 = -2(c - 3)^-3 * (4 - 1)
3/4 = -2(c - 3)^-3 * 3
Dividing both sides by -18:
-1/24 = (c - 3)^-3
At this point, we can see that no matter what value of c we choose within the interval (1, 4), (c - 3)^-3 will always be positive. This means that there is no value of c that satisfies the equation above.
Now, let's consider why this situation does not contradict the mean value theorem (MVT). The MVT states that if a function is continuous on a closed interval [a, b] and differentiable on the open interval (a, b), then there exists at least one value c in (a, b) such that f'(c) = (f(b) - f(a))/(b - a).
In our case, the interval of interest is (1, 4), which satisfies the conditions of the MVT. However, the MVT only guarantees the existence of a value c that satisfies the equation f'(c) = (f(b) - f(a))/(b - a), without specifying a particular form for c.
In our scenario, we found that for the given function f(x) = (x - 3)^-2, there is no value of c that satisfies the equation f(4) - f(1) = f'(c)(4 - 1). But this does not contradict the MVT because the MVT does not guarantee a specific value of c. It only ensures the existence of a value that satisfies the equation. In this case, no value of c satisfies the equation, so the MVT is not contradicted.