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Critical Points

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- 1. TARUN GEHLOT (B.E, CIVIL, HONOURS) Critical Points We will discuss the occurrence of local maxima and local minima of a function. In fact, these points are crucial to many questions related to optimization problems. We will discuss these problems in later pages. Definition. A function f(x) is said to have a local maximum at c iff there exists an interval I around c such that Analogously, f(x) is said to have a local minimum at c iff there exists an interval I around c such that A local extremum is a local maximum or a local minimum. Using the definition of the derivative, we can easily show that: If f(x) has a local extremum at c, then either These points are called critical points.
- 2. TARUN GEHLOT (B.E, CIVIL, HONOURS) Example. Consider the function f(x) = x3 . Then f'(0) = 0 but 0 is not a local extremum. Indeed, if x < 0, then f(x) < f(0) and if x > 0, then f(x) > f(0). Therefore the conditions do not imply in general that c is a local extremum. So a local extremum must occur at a critical point, but the converse may not be true.Example. Let us find the critical points of f(x) = |x2 -x| Answer. We have Clearly we have Clearly we have
- 3. TARUN GEHLOT (B.E, CIVIL, HONOURS) Also one may easily show that f'(0) and f'(1) do not exist. Therefore the critical points are Let c be a critical point for f(x). Assume that there exists an interval I around c, that is c is an interior point of I, such that f(x) is increasing to the left of c and decreasing to the right, then c is a local maximum. This implies that if for (x close to c), and for (x close to c), then c is a local maximum. Note that similarly if for (x close to c), and for (x close toc), then c is a local minimum. So we have the following result: First Derivative Test. If c is a critical point for f(x), such that f '(x) changes its sign as x crosses from the left to the right of c, then c is a local extremum. Example. Find the local extrema of f(x) = |x2 -x| Answer. Since the local extrema are critical points, then from the above discussion, the local extrema, if they exist, are among the points Recall that (1)
- 4. TARUN GEHLOT (B.E, CIVIL, HONOURS) For x = 1/2, we have So the critical point is a local maximum. (2) For x = 0, we have So the critical point 0 is a local minimum. (3) For x = 1, we have So the critical point -1 is a local minimum.
- 5. TARUN GEHLOT (B.E, CIVIL, HONOURS) Let c be a critical point for f(x) such that f'(c) =0. (i) If f''(c) > 0, then f'(x) is increasing in an interval around c. Since f'(c) =0, then f'(x) must be negative to the left of c and positive to the right of c. Therefore, c is a local minimum. (ii) If f''(c) < 0, then f'(x) is decreasing in an interval around c. Since f'(c) =0, then f'(x) must be positive to the left of c and negative to the right of c. Therefore, c is a local maximum. This test is known as the Second-Derivative Test. Example. Find the local extrema of f(x) = x5 - 5 x. Answer. First let us find the critical points. Since f(x) is a polynomial function, then f(x) is continuous and differentiable everywhere. So the critical points are the roots of the equation f'(x) = 0, that is 5x4 - 5 = 0, or equivalently x4 - 1 =0. Since x4 - 1 = (x- 1)(x+1)(x2 +1), then the critical points are 1 and -1. Since f''(x) = 20 x3 , then The second-derivative test implies that x=1 is a local minimum and x= -1 is a local maximum.

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