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

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In summary, the range of the function $y=x+\frac{1}{x^2}$ is all real numbers. This can be proven mathematically by analyzing the derivative and limits of the function at different intervals.
  • #1
leprofece
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\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
 
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  • #2
leprofece said:
\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
 
  • #3
topsquark said:
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
 
  • #4
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.
 
  • #5
leprofece said:
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.
 
  • #6
Rido12 said:
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
 
  • #7
leprofece said:
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:

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

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

What does "Range of $\frac{1}{x} + x = 1/x^2 + x$ is all R" mean?

The statement means that the range of values for the equation $\frac{1}{x} + x = 1/x^2 + x$ is all real numbers (represented by "R"). In other words, all real numbers can be solutions for this equation.

How is the range of this equation determined?

The range of an equation is determined by finding all the possible output values (also known as the dependent variable) that can result from the given input values (or independent variable). In this case, the range is determined by solving the equation for all possible values of x and identifying the corresponding values of y, which make up the range.

Why is it important to know the range of an equation?

Knowing the range of an equation is important because it helps us understand the behavior and characteristics of the equation. It also allows us to determine the domain of the equation, which is the set of all possible input values. Additionally, knowing the range can help us identify any restrictions or conditions that may apply to the equation.

What is the significance of having a range of all real numbers?

Having a range of all real numbers indicates that the equation has no restrictions or limitations on its solutions. This means that any real number can be substituted for x in the equation, and it will result in a valid solution. In other words, the equation is defined for all real numbers, and there are no values that would make it undefined.

Can the range of this equation ever change?

The range of an equation can change if the equation is modified or if certain conditions are applied to it (such as restrictions on the input values). However, for the given equation $\frac{1}{x} + x = 1/x^2 + x$, the range will always be all real numbers, regardless of any modifications or conditions.

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