Amiinprud

The function f(x) = x3 is increasing between 0 and 1 Therefore the supremum of the values on an interval (x i 1;x i) is f(x i) = x3i, and the in mum is f(x i 1) = x3i 1 Thus we can calculate the lower and upper sums of fwith respect to D n L(f;D n) = i=1 (i 1 n)3 1 n = 1 n4 i=1 (i 1)3 = 1 4n4 (n4 2n3 n2) n3 n4 = 1 4 1 2n 1 4n2 ASolution Let f(x) = d(x;y) By Theorem 403, the function f is continuous By Corollary 427, there exists a point a2Xsuch that d(a;y) d(x;y) for all x2X To see that the conclusion may fail if \compact" is replaced by \closed," let M= l1 Let (k) 2l1 be given by (k) n = (1 if n= k 0 if n6= k;Your input find the average rate of change of $$$ f\left(x\right)=x^{2} $$$ on the interval $$$ \left1,3\right $$$ The average rate of change of $$$ f\left(x\right) $$$ on the interval $$$ a,b $$$ is $$$ \frac{f(b)f(a)}{ba} $$$ We have that $$$ a=1 $$$, $$$ b=3 $$$, $$$ f\left(x\right)=x^{2} $$$ Texts For The Corpus Of Nko Collection Conversion And Open Issues ƒx[ƒXŒ^