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Prove that /2 then aota for some x in (0, 1). Prove that the polynomial function

ID: 3147502 • Letter: P

Question

Prove that /2 then aota for some x in (0, 1). Prove that the polynomial function f"(x)=x3-3x + m never has two roots in [O. II, no matter what m may be. (This is an easy consequence of Rolle's 39. Theorem. It is instructive, after giving an analytic proof, to graph fo and f2, and consider where the graph of Im lies in relation to them.) Suppose that f is continuous and differentiable on [0, 1], that f(x) is in [0, l] for each x, and that f(x) 1 for all x in [0, i]. Show that there is exactly one number x in [0, 1] such that f(x) = x. (Half of this problem has been done already, in Problem 7-11.) 40. Prove that the function f (x) = x2-cos x satisfies f(x) two numbers x. 41. (a) 0 for precisely Prove the same for the function f(x) = x-x sin x-cos x. (b) "(c) Prove this also for the function f(x) 2x2-xsin cos2x. (Some preliminary estimates will be useful to restrict the possible location of the zeros of f.) *42. (a) Prove that if is a twice differentiable function with f() = 0 and f(1) = 1 and f,() = f'(1) = 0, then If"(x)2 4 for some x in (0,1). In more picturesque terms: A particle which travels a unit distance in a unit time, and starts and ends with velocity 0, has at some time an acceleration or else f"(x) 3 -4 for some x in . 1). 4. Hint: Prove that either f"(x) 4 for some x in (0, (b) Show that in fact we must have lf"x)>4 for some x in (0, 1) Suppose that f is a function such that f(x) = 1/x for all x > 0 and f(1) 0. Prove that f(xy) = f(x) + f(y) for all x, y > 0, Hint: Find g(x) when g(x) = f(xy). 43. 44. Suppose that f satisfies f" (x) + f' (x)g(x)-f(x) = 0 for some function o Prove that ii Oot t

Explanation / Answer

40

LEt,

g(x)=f(x)-x

f is continuous and hence, g is continuous

Let there be two distinct x,y in [0,1] so that

g(x)=0=g(y)

Without loss of generality we assume, x<y

By Rolle's theorem

there is a c in (x,y) so that

g'(c)=f'(c)-1 =0 which is not possible as f'(c) is not equal to 1 on the interval

So,g(x)=0 for atmost one x

Now we show there is one such x in [0,1] so that g(x)=0

If, g(0)=0 or g(1)=0 we are done

Else

g(0)=f(0)>0

g(1)=f(1)-1<0

By Intermediate value theorem

there is some c in (0,1) so that g(c)=0

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