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4.3.1-eg-1

# 4.3.1-eg-1 - example if we pick the the point-4 in the...

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Example A particular function f has its derivative given by f 0 ( x ) = x 3 + x 2 - 6 x. Find the critical points of f and use them to determine the open intervals where f is increasing and/or decreasing, and to find the x -values where f has relative maxima or minima. Solution: The critical points of f are those points x in the domain of f for which f 0 ( x ) = 0 or f 0 is undefined. In this case, f 0 is defined everywhere, but f 0 ( x ) = x ( x 2 + x - 6) = x ( x + 3)( x - 2) = 0 at the critical points x = - 3 , 0 , 2 . (1) These critical points divide the real line into the open intervals ( -∞ , - 3), ( - 3 , 0), (0 , 2), and (2 , ). To determine whether f is increasing or decreasing on these intervals we make a sign chart with these intervals along the first column, and the factors in f 0 along the first row: interval x ( x + 3) ( x - 2) f 0 ( x ) = x ( x - 3)( x + 2) f ( -∞ , - 3) - - - - decreasing ( - 3 , 0) - + - + increasing (0 , 2) + + - - decreasing (2 , ) + + + + increasing To determine the “+” or “ - ” entries in the columns, we need only compute the given factor at a single point in the designated interval (due to the Intermediate Value Theorem).
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Unformatted text preview: example, if we pick the the point-4 in the interval (-∞ ,-3), then because x < 0 at this point a “-” goes in the column under x ; the factor x + 3 < 0 at the point x =-4, so a “-” goes in the column under ( x + 3); and so on. After determining the sign of f on the given interval, it follows that f is increasing if f > 0 on that interval; f is decreasing if f < 0 on that interval. Finally, candidates for relative maxima/minima are the critical points given in (1): • Because f changes from decreasing to increasing at the point x =-3, f has a relative minimum at x =-3. • The function f changes from increasing to decreasing at the point x = 0 so f has a relative maximum at x = 0. • Finally, f changes from decreasing to increasing at the point x = 2 so f has a relative minimum at x = 2....
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