Let
f(x) = (x − 3)−2.
Find all values of c in (2, 5) such that
f(5) − f(2) = f'(c)(5 − 2).
(Enter your answers as a comma-separated list. If an answer does not exist, enter DNE.)
C=?
If you mean f(x) = (x-3)^-2,
then f is discontinuous at x=3 and the mean value theorem does not apply.
Why did the function go to therapy? It had an identity crisis!
Let's find those values of c, shall we?
First, let's find f(5) and f(2):
f(5) = (5 - 3)^-2 = 1/4
f(2) = (2 - 3)^-2 = 1
Next, let's find f'(x):
f(x) = (x - 3)^-2
f'(x) = -2(x - 3)^-3
Now, let's plug those values into the equation:
1/4 - 1 = f'(c)(3)
-3/4 = f'(c)
Let's solve for c:
-3/4 = -2(c - 3)^-3
Hmm, it seems like we've hit a roadblock here. There doesn't appear to be any values of c in the interval (2, 5) that satisfy this equation. Therefore, the answer is DNE.
Don't worry though, maybe next time c will show up fashionably late to the answer party!
To find the values of c in the interval (2, 5) that satisfy the equation f(5) - f(2) = f'(c)(5 - 2), let's first calculate the values involved.
Given the function f(x) = (x - 3)^(-2), we need to find f(5), f(2), and f'(c).
First, let's find f(5):
f(5) = (5 - 3)^(-2)
= 2^(-2)
= 1/4
Next, let's find f(2):
f(2) = (2 - 3)^(-2)
= (-1)^(-2)
= 1
Now, let's find f'(c):
f'(x) = d/dx[(x - 3)^(-2)]
= -2(x - 3)^(-3) * 1
= -2/(x - 3)^3
Substituting x with c in f'(x), we have:
f'(c) = -2/(c - 3)^3
Now, let's substitute the values into the equation:
f(5) - f(2) = f'(c)(5 - 2)
1/4 - 1 = -2/(c - 3)^3 * 3
-3/4 = -6/(c - 3)^3
Multiply both sides by -4 to eliminate the fraction:
3 = 24/(c - 3)^3
To further simplify, multiply both sides by (c - 3)^3:
3(c - 3)^3 = 24
Divide both sides by 3:
(c - 3)^3 = 8
Taking the cube root of both sides:
c - 3 = 2
Adding 3 to both sides:
c = 5
Therefore, the only value of c in (2, 5) that satisfies the equation is c = 5.
The value of c that satisfies the equation is 5.
To find the values of c in the interval (2, 5) that satisfy the equation f(5) - f(2) = f'(c)(5 - 2), we need to apply the Mean Value Theorem (MVT).
First, let's find the values of f(5) and f(2):
f(x) = (x − 3)^(-2)
f(5) = (5 − 3)^(-2) = 1/4
f(2) = (2 − 3)^(-2) = 1
Now, let's find f'(c). We'll start by finding f'(x):
f(x) = (x − 3)^(-2)
Differentiating both sides using the chain rule, we get:
f'(x) = -2(x - 3)^(-3) * 1
Now, substituting x = c:
f'(c) = -2(c - 3)^(-3)
Now, let's substitute these values back into the MVT equation:
1/4 - 1 = -2(c - 3)^(-3)(5 - 2)
-3/4 = -6(c - 3)^(-3)
(c - 3)^(-3) = 1/2
Taking the cube root of both sides, we have:
c - 3 = (1/2)^(1/3)
c - 3 = 1/2^(1/3)
c = 1/2^(1/3) + 3
So the value of c in the interval (2, 5) that satisfies the equation is:
c = 1/2^(1/3) + 3
Therefore, the value of c is approximately 4.079.