Respuesta :
Answer:
Step-by-step explanation:
This is the Mean Value Theorem for Calculus. The formula for that, at least the one I use for it, resembles yours, but I don't simplify it down quite as far, as I find it easier to work with:
[tex]f'(c)=\frac{f(b)-f(a)}{b-a}[/tex]
That says that the derivative of the function at some value of c within a < c < b
is equal to the slope of the function on the interval.
We need f'(c), f(b) and f(a) to solve this. First the derivative. I have to say that I am assuming the function is
[tex]f(x)=(x-3)^{-2}[/tex]
If so, the derivaitve of this function is
[tex]f'(x)=\frac{-2}{x-3}[/tex]
Next, f(b). We plug in a 5 for x in the function to get that f(b) = .25
Next, f(a). We plug in a 2 for x in the function to get that f(a) = 1
Filling in the MVT:
[tex]\frac{-2}{x-3}=\frac{.25-1}{5-2}[/tex]
Cross multiply to get
-.75(x - 3) = -6 and
-.75x + 2.25 = -6 so
-.75x = -8.25 and
x = 11 (that is the same as c = 11). Now we can find the y value that corresponds to that x value by subbing 11 into x in the function:
[tex]f(11)=(11-3)^{-2}[/tex]
[tex]f(11)=\frac{1}{64}[/tex]
Therefore, the value of c where the slope of the function is the same as the slope of the derivative is at (11, 1/64)
The answer does not exist.
Note - The statement has typing mistakes. Correct form is presented below:
Let [tex]f(x) = (x-3)^{-2}[/tex]. Find all values of [tex]c[/tex] in (2, 5) such that [tex]f(5) - f(2) = f'(c) \cdot (5-2)[/tex]. (Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
In this question we should use the Mean Value Theorem, which states that given a secant line between points A and B, there is at least a point C that belongs to the curve whose derivative exists.
We begin by calculating [tex]f(2)[/tex] and [tex]f(5)[/tex]:
[tex]f(2) = (2-3)^{-2}[/tex]
[tex]f(2) = 1[/tex]
[tex]f(5) = (5-3)^{-2}[/tex]
[tex]f(5) = 1[/tex]
And the slope of the derivative is:
[tex]f'(c) = \frac{f(5) - f(2)}{5-2}[/tex]
[tex]f'(c) = 0[/tex]
Now we find the derivative of the function:
[tex]f'(x) = -2\cdot (x-3)^{-3}[/tex]
[tex]-2\cdot (x-3)^{-3} = 0[/tex]
[tex]-2 = 0[/tex] (ABSURD)
Hence, we conclude that the answer does not exist.
We kindly invite to see this question on Mean Value Theorem: https://brainly.com/question/3957181
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