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.
Let me know if you don’t understand anything. Hope this helped.
Answer:

Step-by-step explanation:
Given:

Solve for:

Solution:
Step 1: Select the formula to solve for 
There are some ways to solve quadratic equations.
One of the simple way (if possible) is performing the factorization.
Another way is using the quadratic formula.
Here, we are going to use factorization.
Step 2: Perform factorization

<=>
<=>
<=>
<=>
<=>
or 
Hope this helps!
:)