45.Because the function is not continuous on [0, 1]. The function does not satisfy the hypotheses of the Mean ValueTheorem, and so it need not satisfy the conclusion of theMean Value Theorem.46.Because the Mean Value Theorem applies to the functiony5sin xon any interval, and y5cos xis the derivative ofsin x. So, between any two zeros of sin x, its derivative,cos x, must be zero at least once.47.f(x) must be zero at least once between aand bby theIntermediate Value Theorem. Now suppose that f(x) is zerotwice between aand b. Then by the Mean Value Theorem,f9(x) would have to be zero at least once between the twozeros of f(x), but this can’t be true since we are given thatf9(x)±0 on this interval. Therefore,f(x) is zero once andonly once between aand b.48.Let f(x)5x413x11. Then f(x) is continuous and differentiable everywhere. f9(x)54x313, which is neverzero between x522 and x1. Since f(22)511 andf(21)1, exercise 47 applies, and f(x) has exactly onezero between x2 and x1.
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This note was uploaded on 10/05/2011 for the course MAC 1147 taught by Professor German during the Fall '08 term at University of Florida.