Ration expressions cause excluded values wherever the denominator equals zero.
So, for any expression like
-5 is an excluded value if
For example, the simplest one would be
In fact, if you try to evaluate this function at -5, you'd have
which is undefined, and thus you can't evaluate the function, and thus -5 is an excluded value.
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;
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.