The answer is b hope that helps


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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:
Step-by-step explanation: (9=4-38-838,256=2-60^7
Answer:
x = 4
Step-by-step explanation:
It is stated that f(x) passes through
(-3, 3), (0, 3), (4, 3)
and g(x) passes through
(-4, -3), (0, 0), (4, 3)
Therefore, both functions pass through (4, 3). That is, the input value x = 4 produces the same output value y = 3 for the two functions.