how do you prove that the unction f(x) = tan(x+x^2)/(1+x+x^2) is continuous for all x=[0,1] except at one point.
i found out that it is not cont at x= 0.84936..but i dont know how to prove that it is continuous now..
The function $\displaystyle f(\theta) = \tan \theta$ has a singularity for $\displaystyle \theta= \frac{\pi}{2}$ so that Your function has a singularity for x satisfying the equation $\displaystyle x^{2}+x-\frac{\pi}{2}=0$. One solution is $\displaystyle \displaystyle x= \frac{-1 + \sqrt{1+2 \pi}}{2} = .849368862...$
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
The function:
$\displaystyle f(x)=\dfrac{\tan (x+x^2)}{1+x+x^2}$
is an elementary function, $\displaystyle 1+x+x^2\neq 0$ for all $\displaystyle x$ real and $\displaystyle \tan \pi/2$ does not exists. Solving $\displaystyle x^2+x=\pi/2$ you'll obtain the point in $\displaystyle [0,1]$ where $\displaystyle f$ is not continuous:
$\displaystyle x=\dfrac{-1+\;\sqrt[]{1+2\pi}}{2}\in{[0,1]}$
Regards.
Edited: Sorry, I didn't see the previous answers.
For example:
(i) $\displaystyle f_1(x)=1+x+x^2$ is continuous in $\displaystyle [0,1]$ (polynomical function).
(ii) $\displaystyle f_2(x)=x+x^2$ is continuous in $\displaystyle [0,1]$ (polynomical function).
(iii) $\displaystyle f_3(x)=\tan x$ is continuous in $\displaystyle \ldots \cup (-\pi/2,\pi/2)\cup\ldots$
(iv) Composition of continuous are continuous.
etc. etc., ...
Regards.