# range

#### jacks

##### Well-known member
Range of $\displaystyle f(x) = x^2+1+\frac{1}{x^2+x+1}\forall x\in \mathbb{R}$

#### chisigma

##### Well-known member
Range of $\displaystyle f(x) = x^2+1+\frac{1}{x^2+x+1}\forall x\in \mathbb{R}$
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...

$\displaystyle 2\ x^{5} +4\ x^{4} +6\ x^{3} +4\ x^{2} -1=0$ (1)

The (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 $[y_{0}, + \infty)$...

Kind regards

$\chi$ $\sigma$

• Sudharaka and jacks

#### checkittwice

##### Member
Range of $\displaystyle f(x) = x^2+1+\frac{1}{x^2+x+1}\forall x\in \mathbb{R}$

jacks,

if a solver can't do this without using calculus, then this problem should not
be posted under "Pre-Calculus." You would post it under "Calculus."

---------- Post added at 11:07 ---------- Previous post was at 10:57 ----------

The (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 $[y_{0}, + \infty)$...
My TI-83 gives rounded values of (which when rounded to 6 decimal places) as

0.379095 and 1.800395, respectively.

• Mr Fantastic

#### jacks

##### Well-known member
Yes checkittwice next time i will consider it

Thanks chisigma

but how can i calculate the real roots of the equation $2x^5+4x^4+6x^3+4x^2-1=0$

#### chisigma

##### Well-known member
Yes checkittwice next time i will consider it

Thanks chisigma

but how can i calculate the real roots of the equation $2x^5+4x^4+6x^3+4x^2-1=0$
As explained in...

... the solutions of the equation $\displaystyle \varphi (x)=0$ is the limit [if limit exists...] of the solution of the difference equation...
$\displaystyle \Delta_{n}= a_{n+1}-a_{n}= f(a_{n})\ ;\ f(x)=- \frac {\varphi(x)}{\varphi^{\ '}(x)}$ (1)
... with a proper 'initial value' $a_{0}$ and the conditions about $a_{0}$ for convergence to the solution are also explained...
$\chi$ $\sigma$