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Calculus I (MAT 145) Dr. Day Thursday March 21, 2013

Calculus I (MAT 145) Dr. Day Thursday March 21, 2013. 4.1: Extreme Values Gateway Quiz #4!. EXTREME VALUE THEOREM.

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Calculus I (MAT 145) Dr. Day Thursday March 21, 2013

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  1. Calculus I (MAT 145)Dr. Day Thursday March 21, 2013 • 4.1: Extreme Values • Gateway Quiz #4! MAT 145

  2. EXTREME VALUE THEOREM • If f is continuous on a closed interval [a, b], then f attains an absolute maximum value f(c) and an absolute minimum value f(d) at some numbers x = cand x = din [a, b]. MAT 145

  3. FERMAT’S THEOREM • If f has a local maximum or minimum at x = c, and if f’(c) exists, then f’(c) = 0. MAT 145

  4. FERMAT’S THEOREM The following examples caution us against reading too much into the theorem. • We can’t expect to locate extreme values simply by setting f’(x) = 0 and solving for x. MAT 145

  5. CRITICAL NUMBERS A critical numberof a function f is a number x = cin the domain of f such that either: f ’(c) = 0 or f’(c) does not exist. MAT 145

  6. CLOSED INTERVAL METHOD To determine the absolute maximum and minimum values of a continuous function f on a closed interval [a, b]: • Find the values of f at the critical numbers of fin (a, b). • Find the values of f at the endpoints of the interval. • The largest value from (1) and (2) is the absolute maximum value. The smallest is the absolute minimum value. MAT 145

  7. CLOSED INTERVAL METHOD Determine the absolute maximum and absolute minimum values of the function:f(x) = x3 – 3x2 + 1 -½ ≤ x ≤ 4 MAT 145

  8. MAXIMUM & MINIMUM VALUES • The Hubble Space Telescope was deployed on April 24, 1990, by the space shuttle Discovery. MAT 145

  9. MAXIMUM & MINIMUM VALUES A model for the velocity of the shuttle during this mission—from liftoff at t = 0 until the solid rocket boosters were jettisoned at t = 126 s—is given by: v(t) = 0.001302t3 – 0.09029t2 + 23.61t – 3.083 (in feet per second) MAT 145

  10. MAXIMUM & MINIMUM VALUES Using this model, estimate the absolute maximum and minimum values of the acceleration of the shuttle between liftoff and the jettisoning of the boosters. v(t) = 0.001302t3 – 0.09029t2 + 23.61t – 3.083 MAT 145

  11. Assignments WebAssign • 4.1 (I) due tonight • 4.1 (II) due tomorrow night MAT 145

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