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.
=8x + (-3) (5) + (-3) (9x)
=8x + -15 + -27x
=8x + -15 + -27x
(8x + -27x) + (-15)
19x +-15
19x -15
Answer:
1.09
Step-by-step explanation:
hope this helps
-mercury
A and d seem correct to me.
b and c are definitely wrong.
Answer:
2/10 + 7/10
9/10
Step-by-step explanation:
I did the test.