Question

Problem 3. Prove Theorem 1 as tollows [Assume all conditions of the Theorem are met. In many parts, it will be useful to consider the sign of the right side of the formula-positive or negative- ad to write the appropriate inequality] (a) Since f"(x) exists on [a, brx) is continuous on [a, b) and differentiable on (a,b), soMean Value Thorem applies to f,on this interval. Apply MVTtof"m[x,y], wherc α zcysb. to show that ry)2 f,(x), İ.e. that f, is increasing on [a, b]. Conclude that f,() > 0 on [a,b] by comparing to f(a). Note: We have used this idea, that if the derivative of a function is positive, the function is increasing before. Mean Value Theorem provides an casy way of proving it. There are other more complicated ways, such as using the definition of the derivative along with "compactness" (b) Since f,(x) exists on a, bİ. f(x) is continuous on [a,b] and differentiable on (a, b), so Mean Value Theorem applies to fon this interval. Apply MVT to f on [x.yl. where a Sx(x)i.e that fis strictly increasing on a,b. Conclude that f(x) >0 on (a, b] by comparing to (a) (c) We now show that ifx, e [a, b], then so iszn+1. and xn.i Xn, (i) First consider xn € (a,b]. We can rearrange the rocursion fomula.,-채-fua) to Use this to prove Sb (ii) Still considering x, E (a, b, apply MVT to fon a, x). Since f" is increasing on a, b. f(c)sf(x) for any e in (a,). Use this to turn the MVT equation into an inequality, and rearrange to show a sX Soa Sx, Sb (iii) Finally, consider xi.-a. Use the recursion formula to show In+1 = a, so Thus, forany x" 티a, bla x,+:S x.s b, which is what we sought. decreasing sequence with values in la, bl.a Xy XS xe b. By induction, {제 is a (d) [No work noodod. I've done it for you] We have showm (xis a decrcasing sequence of real numbers, bounded below by a. A thoorem in real analysis proves that the soquence converges to some real mumber L (which must be in la,b, in this cincumstance), so lim-L for some L la.b we show that the sequence goes all the way down to a, ic. 1.a. Taking thclimit as n →ue on both sides of the recursion formula, and using limit laws, the continuity of f and fon [a,b], and the fact So (xn-o decreases to a as-. oo; Theorem l is proved. Theorem 1 . Suppose f has a zero at a (so f(a) = 0), f, (a) > 0, and f" (x) 0 on some interval a, b] with b > a. [This is the case in the picture shown earlier.] Then for any xo E [a, b], the xo-xo sequencexnico defined recursively by xnx n2 o decreases to a as n- decreases to a as n → oo. Tem, n2 o r'an)

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