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It is clear that $\displaystyle \lim_{x \rightarrow + \infty} f(x) = \lim_{x \rightarrow - \infty} f(x)= + \infty$ so that, the range is defined if we find the absolute minimum of f(x). Proceeding with the standard method the values of x where $f^{\ '}(x)$ vanishes are the root of the equation...Range of $\displaystyle f(x) = x^2+1+\frac{1}{x^2+x+1}\forall x\in \mathbb{R}$
Range of $\displaystyle f(x) = x^2+1+\frac{1}{x^2+x+1}\forall x\in \mathbb{R}$
My TI-83 gives rounded values of (which when rounded to 6 decimal places) asThe (1) has a single real root in $x_{0} \sim > > .379093$, < < so that the minimum is $y_{0}=f(x_{0}) \sim > > 1.800394$ < <
and the range of f(*) is $[y_{0}, + \infty)$...
As explained in...Yes checkittwice next time i will consider it
Thanks chisigma
but how can i calculate the real roots of the equation $2x^5+4x^4+6x^3+4x^2-1=0$