Suppose that some value, c, is a point of a local minimum point.
The theorem states that if a function f is differentiable at a point c of local extremum, then f'(c) = 0.
This implies that the function f is continuous over the given interval. So there must be some value h such that f(c + h) - f(c) >= 0, where h is some infinitesimally small quantity.
As h approaches 0 from the negative side, then:

As h approaches 0 from the positive side, then:

Thus, f'(c) = 0
Answer:
x=3/7
Step-by-step explanation:
6-2x=3+5x
subtract 5x from both sides
6-2x-5x=3
subtract 6 from both sides
-2x-5x=3-6
add like terms
-7x=-3
divide both sides by -7
x=3/7
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