• Jul 30th 2012, 04:35 AM
Atena369
Determine the range for $\displaystyle f(x) = \frac{x}{{{x^2} + 1}}.$ I'll be grateful if someone can explain me how to solve this kind of problem.
• Jul 30th 2012, 05:15 AM
emakarov
Re: Quadratic function and its range
I find the range of a function by analyzing its behavior: by finding the domain, maximums and minimums, periods of increase/decrease, asymptotes, behavior at plus/minus infinity and by drawing the graph. I may be tricky to find the extremums of this function without finding the derivative.
• Jul 30th 2012, 07:12 AM
Atena369
Re: Quadratic function and its range
I know how to find the range of a function; the tricky thing (for me) is that this function is a fraction, therefore I don't know how to calculate delta and I have no idea which are coefficients. In regards to find the derivative, I haven't studied that yet...
• Jul 30th 2012, 07:15 AM
emakarov
Re: Quadratic function and its range
Quote:

Originally Posted by Atena369
I don't know how to calculate delta and I have no idea which are coefficients.

What do you mean by delta and coefficients?
• Jul 30th 2012, 07:20 AM
Atena369
Re: Quadratic function and its range
Quote:

Originally Posted by emakarov
What do you mean by delta and coefficients?

A quadratic function is in this form: f(x)=a(x^2) + bx+ c, where a,b,c are real numbers and (in my country at least) are called coefficients. delta=b^2 - 4*a*c
Sorry if my expression is confusing; I haven't been learning maths in English.
• Jul 30th 2012, 07:37 AM
emakarov
Re: Quadratic function and its range
Draw the graph of x and $\displaystyle 1/(x^2 + 1)$. You may draw the second graph approximately, by calculating the value of the function at several points and connecting them in a smooth way. Now draw (approximately) the graph of what you think the product of these two functions would be.

Note the following things.

f(0) = 0.

f(x) is odd. What does this mean for the graph of f(x)?

When x is close to 0, $\displaystyle x^2$ is small and so $\displaystyle 1/(x^2 + 1)$ is approximately 1. Therefore, around 0 we have f(x) ≈ x. In particular, f(x) increases for small x > 0.

For large positive x, $\displaystyle x^2 + 1$ far exceeds x. In fact, 1 is very small compared to $\displaystyle x^2$, so f(x) ≈ x / $\displaystyle x^2$ = 1 / x. The important thing is that f(x) decreases for large x > 0 and tends to 0.

This means that f(x) must have a maximum, possibly several, when x > 0. Let's assume that there is only one maximum, i.e., that f(x) first increases and then decreases when x > 0. Then the value of f(x) at the point of maximum would be the upper bound of the range of f(x).

To find the maximum, you can do the following trick. Consider the horizontal line (which is the graph of) y = y₀ for some y₀. If y₀ is greater than the maximum of f(x), the line has no intersections with the graph of f(x). If y₀ is less than the maximum, there are two intersections. When y₀ equals the maximum, the horizontal line just touches the graph and there is one intersection point. Let's consider the equation $\displaystyle x / (x^2 + 1) = y_0$, i.e., $\displaystyle y_0x^2 - x + y = 0$. This is a quadratic equation in x, so you can calculate the discriminant and find for which y₀ the equation has one root. This y₀ is the maximum of f(x).