Find the range for f(x) = x^2 1, for x < 0. y ≥ 1 y > 1 y < 1 y ≤ 1
1 answer:
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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
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: 14.96ft
Step-by-step explanation:
1. To solve this exercise you must apply the Pythagorean Theorem, which is:
Where is the hypotenuse, and
and
are the other sides of the triangle.
2. You have that the sides given measures 12 inches. 1 feet has 12 inches. Therefore, this side measures 1 feet.
2. Then, when you solve for one of the sides and substitute the values given in the problem into the formula shown above, you obtain that the length of the side of the rigth triangle is: