Dillard

2021-08-21

The properties of function $f\left(x\right)$ as
1) The domain of $f$ is $\left(-\mathrm{\infty },\mathrm{\infty }\right)$
2) $f\left(3\right\}=7$
3) $f$ is continuous st $x=2$
4) $f$ is not continuous at $x=4$ and
5) $\underset{x\to +\mathrm{\infty }}{lim}f\left(x\right)=1$
To find: The equation of function $f\left(x\right)$ which satisfies all given properties.

stuth1

Step 1
Explanation:
Consider the function

The above function is defined over all real number therefore the domain of the function is $\left(-\mathrm{\infty },\mathrm{\infty }\right)$,
Hence the property (1) is satisfied.
Evaluating $f\left(3\right)$,
$f\left(3\right)=\frac{3\left(3-3\right)+70}{{3}^{2}+1}=\frac{70}{10}=7$
Hence the property (2) is satisfied.
Step 2
Evaluating $\underset{x\to {2}^{+}}{lim}f\left(x\right),$
$\underset{x\to {2}^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f\left(2+h\right)$
$⇒\underset{x\to {2}^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{\left(2+h\right)\left(2+h-3\right)+70}{{\left(2+h\right)}^{2}+1}$
$⇒\underset{x\to {2}^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{\left(2+h\right)\left(h-1\right)+70}{{\left(2+h\right)}^{2}+1}$
$⇒\underset{x\to {2}^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{{h}^{2}+h+68}{{h}^{2}+4h+5}$
$⇒\underset{x\to {2}^{+}}{lim}f\left(x\right)=\frac{68}{5}$
Evaluating $\underset{x\to {2}^{-}}{lim}f\left(x\right),$
$\underset{x\to {2}^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f\left(2-h\right)$
$⇒\underset{x-{2}^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{\left(2-h\right)\left(2-h-3\right)+70}{{\left(2+h\right)}^{2}+1}$
$⇒\underset{x-{2}^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}$

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