MATH 140 Lecture Notes - Lecture 26: Mean Value Theorem, Inflection Point, Inflection

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Review problems of f. show work: 1b. Suppose that g is continuous on [0, 3] and differentiable on (0, 3). Also assume that g has 5 zeroes on [0, 3]. What is the minimum number of critical numbers in the interval [0, 3]. Let (cid:1859) (cid:4666)(cid:1872)(cid:4667)=(cid:884)(cid:1872)(cid:3117)(cid:3119)+sec(cid:1872)tan(cid:1872)+(cid:888)(cid:1857) (cid:2870) for all real t. assume that g(0) = 3. Find a: (cid:1859)(cid:4666)(cid:1872)(cid:4667)=(cid:884)(cid:4672)(cid:2871)(cid:2872)(cid:4673)(cid:1872)(cid:3120)(cid:3119)+sec(cid:1872)+(cid:888)(cid:4672) (cid:2869)(cid:2870)(cid:4673)(cid:1857) (cid:2870)+=(cid:2871)(cid:2870)(cid:1872)(cid:3120)(cid:3119)+sec(cid:1872) (cid:885)(cid:1857) (cid:2870), (cid:1859)(cid:4666)(cid:882)(cid:4667)=(cid:885)=(cid:882)+sec(cid:882) (cid:885)+=(cid:883) (cid:885)+= (cid:884)+ =(cid:887) formula for g(t) for all real t, 2a. State carefully the mean value theorem: let f(x) be continuous on an interval [a, b] and differentiable on (a, b); then, (cid:1859)(cid:4666)(cid:1872)(cid:4667)=(cid:2871)(cid:2870)(cid:1872)(cid:3120)(cid:3119)+sec(cid:1872) (cid:885)(cid:1857) (cid:2870), 3a. Find all horizontal asymptotes of h. show work. lim (cid:2871) (cid:3032) (cid:3118) (cid:3032) (cid:3118) = (cid:883) (cid:2869)+(cid:3032) (cid:3118)= lim (cid:3032) (cid:3118) lim (cid:2871) (cid:3032) (cid:3118) (cid:2869)+(cid:3032) (cid:3118)=lim (cid:2871)(cid:2869)=(cid:885: thus, horizontal asymptotes are y = -1 and y = 3, 3b.

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