Given f(x) = (x - 4)^2 - 1, what is the change made to g(x) if the parent function is f(x) = x^2?
1 answer:
You might be interested in
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
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