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Critical points

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Critical Points

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Critical points

  1. 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. 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. 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. 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. 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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