Range and obliquous asymptotes

[leprofece]

\(\displaystyle \frac{1}{x}+x\)

as i canot see it the function is x +1/x
Find the range and the obliquous asymptote

Ok it looks to be easy but the most of books say that I must see the graph but teacher asks me in a math way
How can I calculate it??
a book tells me that I must solving for x

about athe oblique asymptote i must use a formula that i forget

The answers are Range x>2 y x<2
asympote in x but in what number ?

 

[evinda]

\(\displaystyle \\frac{1}{x}\)+ x

as i canot see it the function is x +1/x
Find the range and the obliquous asymptote

Ok it looks to be easy but the most of books say that I must see the graph but teacher asks me in a math way
How can I calculate it??
a book tells me that I must solving for x

about athe oblique asymptote i must use a formula that i forget

The answers are Range x>2 y x<2
asympote in x but in what number ?

$$f(x)=x+\frac{1}{x}=\frac{x^2+1}{x}$$

We notice that the polynomial in the numerator is a higher degree $(2^{nd})$ than the denominator $(1^{st})$, so we know that we have a slant asymptote.

$$f(x)=x+\frac{1}{x}$$

When $x$ gets large, $\frac{1}{x}$ gets very small,so the function $f$ will get close to $x$.Therefore,the asymptote is parallel to $y=x$.

To find the actual asymptote,we look at the limit of $f(x)-x:$

$$\lim_{x \to +\infty} \left ( x+\frac{1}{x}-x \right )=\lim_{x \to +\infty} \frac{1}{x}=0$$

The limit is $0$,so for large $x$, $f(x)$ is close to $x$,which is the asymptote.

Otherwise, you can just use polynomial division.
$f(x)=\frac{x^2+1}{x}=x+\frac{1}{x}$
So,we see that the function has been rewritten after long division has been performed.

The last term of $f$ becomes negligible for large $x$, so the asymptote is $x$ .

 

[leprofece]

Ok thanks but In what value
in any x value
in 0
in x +
or in x -

And I Need the range
it must be calculate in math way without seeing the graph

 

[evinda]

Ok thanks but In what value
in any x value
in 0
in x +
or in x -

And I Need the range
it must be calculate in math way without seeing the graph

$$f(x)=x+\frac{1}{x}$$

The domain of $f$ is $\{ x \in \mathbb{R}: x \neq 0 \}$

$$f'(x)=1-\frac{1}{x^2}$$

$$f'(x)=0 \Rightarrow x^2=1 \Rightarrow x=\pm 1$$

We can see that $f$ is increasing at the interval $(-\infty,-1]$, decreasing at the intervals $[-1,0)$ and $(0,1]$ and increasing at $[1,+\infty)$

The range of the interval $(-\infty,-1]$ is $R_1=(\lim_{x \to -\infty} f(x),f(-1)]=(-\infty,-2]$.

The range of the interval $[-1,0)$ is $R_2=(\lim_{x \to 0} f(x),f(-1)]=(-\infty,-2]$.

The range of the interval $(0,1]$ is $R_3=[f(1), \lim_{x \to 0} f(x))=[2,+\infty)$

The range of the interval $[1,+\infty)$ is $R_4=[f(1),\lim_{x \to +\infty} f(x))=[2,+\infty)$

Therefore,the range of the function is: $R=R_1 \cup R_2 \cup R_3 \cup R_4=(-\infty,-2] \cup [-2,+\infty )$.

 

[leprofece]

Thanks And what is the value of the asymptote?

 

[topsquark]

evinda gave you that in post #2. The asymptote is a line, not a value.

-Dan

 

[MarkFL]

Another method you could use to find the range is to write the equation as:

\(\displaystyle x^2-yx+1=0\)

Require the discriminant to be non-negative:

\(\displaystyle y^2-4\ge0\)

Hence:

\(\displaystyle |y|\ge2\)