Answer:
This contradicts the Mean Value Theorem since there exists a c on (1, 7) such that f '(c) = f(7) − f(1) (7 − 1) , but f is not continuous at x = 3
Step-by-step explanation:
The given function is

When we differentiate this function with respect to x, we get;

We want to find all values of c in (1,7) such that f(7) − f(1) = f '(c)(7 − 1)
This implies that;




![c-3=\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c-3%3D%5Csqrt%5B3%5D%7B63.15789%7D)
![c=3+\sqrt[3]{63.15789}](https://tex.z-dn.net/?f=c%3D3%2B%5Csqrt%5B3%5D%7B63.15789%7D)

If this function satisfies the Mean Value Theorem, then f must be continuous on [1,7] and differentiable on (1,7).
But f is not continuous at x=3, hence this hypothesis of the Mean Value Theorem is contradicted.
Answer:
Step-by-step explanation:
12:32 is simply 3:8 but divided by 4 to both sides.
Answer:
-2 5/9
Step-by-step explanation:
w + (-x) - 2/3
Let w = -5/9 and x = 4/3
-5/9 - 4/3 - 2/3
Get a common denominator
-5/9 -4/3*3/3 -2/3*3/3
-5/9 -12/9 -6/9
Add
-23/9
Change to a mixed number
-2 5/9
Answer:
yes that is the correct answer