15) Give an example of a function defined on (a,b) that is not continuous on (a,b) but attains both relative and absolute extrema.

4) prove that between two consecutives roots f(x)=1–e^x sin x there exists at least one root of g(x) = 1 + e^x cos x.

Glue a funct (a,6) that is on ex cmp la o defined on in d chains absolut c extrema 4) pr two there ex H at least one rout e (OS X


xxx extreme xxxxx xxxxxxx xxxxxx that xxxxxxxxxy on x xxxxxx xxxxxxxx is xxxxxxxxxx to xxxxxx xxxx xxx function xxxxxxx a xxxxxxx xxx xxxxxxx. However, xxxx condition xx xxx xxxxxxxxy. Consider

x(x) = {x, xxx x=x;

         {x, elsewhere.

xxxxxxy, xxx(x)=x, xxx(x)=x, but x is xxxxxxxxxxxxx xx x=x.


xxxx-xxxxxxxxxx functions xxxx xxxx xxxxxxxy.

f(x) = 1 - x^x * sin(x)
x(x) = x xxxx xxx(x) = x^(-x)
xxx xx xx plot xxx(x) and x^(-x), xx xxx that xxx curves xxxxxxxxx xx xxxx positive xxxxx of xxx(x). x.x. xxx first xxxxxxxxxxxx is xx x < x < /2 xxx xxx xxxxxx at /x < x < . Consecutive xxxxxxxxxxxxx are xx xx < x_i < 2n + /x xxx 2n + /2 < x_x < (2n+1)
xxxxx x_i xxx x_x xxx consecutive xxxxx of x(x). xx the xxxxx of x(x), xxx(x) > 0 xxx cos(x_i) > x xxxxx cos(x_j) < 0
[xxxxxxx xx xxx quadrants xx x_i xxx x_x] Now xxxxx sin(x) = x^(-x) xxx(x) = ± [1 - x^(-xx)] [xxx sign xxxxxxx on xxx xxxxxxxx] g(x) = 1 + x^x * cos(x)
= 1 ± (x^(xx) - 1)
At x_x, xxx(x_x) > 0 xxxxxxxxx g(x_i) = x + (e^(2x_i) - 1) > x At x_x, cos(x_j) < x, xxxxxxxxx g(x_j) = 1 - (x^(xx_x) - 1)
xxxxx x_j > /x, x^(x*x_x) > x, thus x(x_x) < x
xxxxx x_i xxx x_x xxx consecutive xxxxx of x(x) xxx x(x_x) and x(x_x) have xxxxxxxx xxxxx, xxx since x(x) is xxxxxxxxxx, xxxxx xxxx be xx least xxx xxxx xx g(x) xx between xxxx.