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:
x=5, y=-3, x=-4, y=9
Step-by-step explanation:
1. x=5, is a vertical line cutting the x axis at 5
2. y=-3 is a horizontal line cutting the y axis at -3
3. x= -4, is a vertical line cutting the x-axis at -4
4. y=9, is a horisontal line cutting the y-axis at 9
Https://us-static.z-dn.net/files/d1e/04b1f4bb9da77e9bfea077c2aeee7a27.jpg
ANSWER
No solution
EXPLANATION
The first equation is

and the second equation is

We equate the two equations to obtain;

This implies that


There is no real number whose square is -1.
Therefore, the equation has no solution.