1873_solutions

1873_solutions

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Unformatted text preview: n x 2 sin then drag each of the expressions different colors. x2 and x2 1 x on the interval . 2, . 2 and into your plot. Revise the plot and give the components 0.04 0.02 -0.2 0 -0.1 0.1 x 0.2 -0.02 -0.04 b. Prove that the function f is differentiable on R but that the function f is not continuous at the number 0. For each t 0 we have t 2 sin 1 ft f0 t   t sin 1 t t t0 and, since ft f0  t sin 1 |t | t t0 for each t 0 we can deduce from the sandwich theorem that ft f0  0. lim t0 t0 Now given any x 0 we have f x  2x sin 1 cos 1 x x and the latter expression does not approach a limit as x 0. Therefore f is not continuous at 0. 5. This exercise concerns the function f defined by the equation fx  x 3 sin 0 239 1 x if x 0 if x  0 . a. Ask Scientific Notebook to make a 2D plot of the expression x 3 sin and then drag each of the expressions different colors.x 3 sin 1 x x3 and x3 1 x on the interval . 05, . 05 into your plot. Revise the plot and give the components 0.0001 5e-05 -0.04 0 -0.02 0.02 x 0.04 -5e-05 -0.0001 b. Prove that the function f is continuous at the number 0 but does not have a derivative there. The fact that f 0  0 follows in exactly the same way as Exercise 4. Now given any x 0 we have f x  3x 2 sin 1 x cos 1 x x and since |f x | 3x 2  |x | for x 0 we have lim f x  0  f 0 . To see why f has no derivative at 0 we can argue as in Exercise 4. x0 6. Suppose that f is a function defined on an open interval a, b and that x a, b . a. Prove that if f x exists then f x  lim h0 Solution: We assume that f f xh h fx . x exists. Thus f x  lim ft tx t fx . x To show that f xh f x , h  0. Choose a number   0 such that the inequality ft fx fx  tx f x  lim h0 suppose that holds whenever t a, b and t x and |t x |  . Then whenever h 0 and |h |   and h a x, b x we deduce from the fact that x  h a, b and | x  h x |   that f xh f x f xh f x fx  fx . h xh x b. Prove that if the limit f xh h exists then f x exists and is equal to this limit. lim h0 c. Prove that if f x exists then 240 fx . f x  lim f xh Hint: Use the fact that whenever h f xh fx 2h h fx h 2h h0 0 and is sufficiently small we have f xh f x fx h fx  2h 1 f xh f x  1 f x h  2 2 h h fx d. Prove that if f x exists then f x  lim lim ux ft tx t fu u Since f x exists, the function f must be continuous at x. Therefore the right side is ft fu ft fx  lim f x. lim lim tu tx tx ux tx Exercises on the Mean Value Theorem 1. a. Given that f is a function defined on an interval S and that f x  0 for every x S, prove that f must be constant on S. Suppose that a and b are numbers in the interval S and a  b. We shall show that f a  f b . Applying the mean value theorem to the function f on the interval a, b we choose a number c between a and b such that fb fa fc . ba Since f c  0 we deduce that f b f a  0 which gives us f a  f b . b. Given that f is a function defined on an interval S and that f x  0 for every x strictly increasing on S. S, prove that f must...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

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