Range of $\frac{1}{x} + x = 1/x^2 + x$ is all R

[leprofece]

\frac{1}{x}+ x = 1/x^2+ x
The answer is all R
In a math way if i have vertical asymptote in 0 how can i explain the range is that

 

[topsquark]

\frac{1}{x}+ x = 1/x^2+ x
The answer is all R
In a math way if i have vertical asymptote in 0 how can i explain the range is that

This is an equation for x, not a function. What are you referring to as the "range?"

-Dan

 

[leprofece]

This is an equation for x, not a function. What are you referring to as the "range?"

-Dan

THEN IS y= x + 1 / x^2
Range of that function

 

[Dethrone]

Assuming you mean:

\(\displaystyle y = x + \frac{1}{x^2}\), then the range is all real numbers. The fact that there is a vertical asymptote doesn't change that fact; it only affects the domain.

 

[MarkFL]

THEN IS y= x + 1 / x^2
Range of that function

Since you have posted this as a correction, which is the same as another thread, I deleted the other thread as a duplicate.

 

[leprofece]

Assuming you mean:

\(\displaystyle y = x + \frac{1}{x^2}\), then the range is all real numbers. The fact that there is a vertical asymptote doesn't change that fact; it only affects the domain.

Oh it doesn't like my teacher he says I need to prove it in a math way unless it is a theoretical matter

 

[evinda]

Oh it doesn't like my teacher he says I need to prove it in a math way unless it is a theoretical matter

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

The domain of $f$ is $(-\infty,0) \cup (0,+\infty)$

$$f'(x)=1-\frac{2x}{x^4}=1-\frac{2}{x^3}=\frac{x^3-2}{x^3}$$

$$f'(x)=0 \Rightarrow x^3=2 \Rightarrow x=\sqrt[3]{2}$$

We can see that $f'(x)>0$ for $x>\sqrt[3]{2}$ and for $x<0$ and $f'(x)<0$ for $0<x<\sqrt[3]{2}$

Therefore, $f$ is increasing at $(-\infty,0)$, decreasing at $(0, \sqrt[3]{2} ]$ and increasing at $[\sqrt[3]{2},+\infty)$

$$R_1=(\lim_{x \to -\infty} f(x), \lim_{x \to 0} f(x))=(-\infty,+\infty)$$

$$R_2= [ f(\sqrt[3]{2}), \lim_{x \to 0} f(x))=[ \frac{3}{2^{\frac{2}{3}}},+\infty)$$

$$R_3=[ f(\sqrt[3]{2}), \lim_{x \to +\infty} f(x))=[ \frac{3}{2^{\frac{2}{3}}}, +\infty)$$

The range of $f$ is:

$$R=R_1 \cup R_1 \cup R_3=(-\infty,+\infty)$$