Answer to Question #164476 in Real Analysis for Sumaira

Question

Construct an example of a function  f(x) that is defined at every point in a closed interval and whose values at the end points have opposite signs but still f(x)=0 has no solution in the interval.

Answer 1

$Solution: \\ Example~1) \\ Let~ f(x)=x^2+sin(x)-1 \\To~apply~the ~Intermediate~Value~Theorem~to~the~equation~f(x)=0~on~the~interval~[0,\pi], we ~evaluate~f(0) ~and~ f(\pi) \\f(0)=0^2 +sin(0)-1=-1<0~and~f(\pi)=\pi^2+sin(\pi)-1=\pi^2-1>0 \\Then~ by ~ Intermediate~Value~Theorem,there ~is ~ a~ solution~of~f(x)=0 \\in~the ~interval~[0,\pi]. \\Thus~ we ~conclude ~that~there~is~a ~root~of~the~function~f(x)=x^2+sin(x)-1~in~the~interval~[0,\pi] \\ \\Example ~2): \\Let~f(x)=\frac{1}{x\sqrt{1-x^2}} defined on -1<x<1 \\Its ~values ~are~negative~for~negative~x,positive~x,but~never ~zero. \\Example~3) : \\Let~f(x)=1~if~-1 \le x \le 0~and f(x)=-1~if ~0<x \le1 \\Here~f(x)~has ~opposite~signs~at~the~end~points~while~it~is~never~zero~in~te~interval.$