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.
Ok so the first one is 225 miles but i can't do the graph
and i'm not great at math so give me a bit
4/8-3/8=1/8more
Therefore Keith at 1/8 more of the sandwich than Henry.
Answer:
11 because as you see 3 and 6 if you divide it the scale factor is 1/2 so you just do 5.5 times 2 and get 11
Step-by-step explanation:
Answer:
2.
Step-by-step explanation:
The first numbers in the ordered pairs are the x values;
Difference = 1 - 3 = -1
The absolute value is 2.