Suppose that f is continuous at a critical point x0. x0 is interior point ofa small interval I. If f ′(x) < 0 for all x ∈ I, x < x0 and f ′(x) > 0 for allx ∈ I, x > x0, then show that f has a relative minimum at x0
Question
Suppose that f is continuous at a critical point x0. x0 is interior point ofa small interval I. If f ′(x) < 0 for all x ∈ I, x < x0 and f ′(x) > 0 for allx ∈ I, x > x0, then show that f has a relative minimum at x0
Solution
To show that f has a relative minimum at x0, we can use the first derivative test.
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First, let's recall the first derivative test. According to the first derivative test, if f'(x) changes sign from negative to positive at x0, then f has a relative minimum at x0.
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Given that f is continuous at x0 and x0 is an interior point of a small interval I, we can conclude that f is defined and continuous on I.
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We are given that f'(x) < 0 for all x ∈ I, x < x0. This means that the function is decreasing on the left side of x0.
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We are also given that f'(x) > 0 for all x ∈ I, x > x0. This means that the function is increasing on the right side of x0.
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Since f is continuous at x0 and f'(x) changes sign from negative to positive at x0, we can conclude that f has a relative minimum at x0.
Therefore, we have shown that if f is continuous at a critical point x0, and f'(x) < 0 for all x ∈ I, x < x0, and f'(x) > 0 for all x ∈ I, x > x0, then f has a relative minimum at x0.
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