Show that the δ 1 of the conclusion is the δ of the

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show that the δ 1 of the conclusion is the δ of the hypotheses, and indeed this is generally untrue. 109. Inverse Functions. Suppose in particular that f ( x, y ) is of the form F ( y ) - x . We then obtain the following theorem. If F ( y ) is a function of y , continuous and steadily increasing ( or decreasing ) , in the stricter sense of § 95 , in the neighbourhood of y = b , and F ( b ) = a , then there is a unique continuous function y = φ ( x ) which is equal to b when x = a and satisfies the equation F ( y ) = x identically in the neighbourhood of x = a . The function thus defined is called the inverse function of F ( y ). Suppose for example that y 3 = x , a = 0, b = 0. Then all the conditions of the theorem are satisfied. The inverse function is x = 3 y . If we had supposed that y 2 = x then the conditions of the theorem would not have been satisfied, for y 2 is not a steadily increasing function of y in any interval which includes y = 0: it decreases when y is negative and increases when y is positive. And in this case the conclusion of the theorem does not hold, for y 2 = x defines two functions of x , viz. y = x and y = - x , both of which vanish when x = 0, and each of which is defined only for positive values of x , so that the equation has sometimes two solutions and sometimes none. The reader should consider the more general equations y 2 n = x, y 2 n +1 = x, in the same way. Another interesting example is given by the equation y 5 - y - x = 0 , already considered in Ex. xiv . 7. Similarly the equation sin y = x
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[V : 109] LIMITS OF FUNCTIONS OF A CONTINUOUS VARIABLE 233 has just one solution which vanishes with x , viz. the value of arc sin x which vanishes with x . There are of course an infinity of solutions, given by the other values of arc sin x (cf. Ex. xv . 10), which do not satisfy this condition. So far we have considered only what happens in the neighbourhood of a particular value of x . Let us suppose now that F ( y ) is positive and steadily in- creasing (or decreasing) throughout an interval [ a, b ]. Given any point ξ of [ a, b ], we can determine an interval i including ξ , and a unique and continuous inverse function φ i ( x ) defined throughout i . From the set I of intervals i we can, in virtue of the Heine-Borel Theorem, pick out a finite sub-set covering up the whole interval [ a, b ]; and it is plain that the finite set of functions φ i ( x ), corresponding to the sub-set of intervals i thus selected, define together a unique inverse function φ ( x ) continuous through- out [ a, b ]. We thus obtain the theorem: if x = F ( y ) , where F ( y ) is continuous and increases steadily and strictly from A to B as x increases from a to b , then there is a unique inverse function y = φ ( x ) which is continuous and increases steadily and strictly from a to b as x increases from A to B . It is worth while to show how this theorem can be obtained directly without the help of the more difficult theorem of § 108 . Suppose that A < ξ < B , and consider the class of values of y such that (i) a < y < b and (ii) F ( y ) 5 ξ . This class has an upper bound η , and plainly F ( η ) 5 ξ . If F ( η ) were less than ξ , we could find a value of y such that y > η and F ( y ) < ξ
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