Question

Find domain and range of (answers should be subsets of R):

f(x)= x/(x^2+1)

Answer 1

ANSWER on Question #76193 - Math - Discrete Mathematics

Find domain and range of (answers should be subsets of $\mathbb{R}$):

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

SOLUTION

$$f(x) = \frac{x}{x^2 + 1} \rightarrow \text{Domain}: x^2 + 1 \neq 0$$ $$x^2 + 1 \neq 0 \rightarrow x^2 \neq -1 \rightarrow x \in \mathbb{R}$$

Conclusion,

$$f(x) = \frac{x}{x^2 + 1} \rightarrow \text{Domain}: x \in \mathbb{R}$$

Consider the range.

The first method

$$x^2 + 1 \geq 2x \text{ because } (x - 1)^2 \geq 0,$$ $$x^2 + 1 \geq -2x \text{ because } (x + 1)^2 \geq 0.$$

Divide inequalities $x^2 + 1 \geq 2x$ and $x^2 + 1 \geq -2x$ by $x^2 + 1 > 0$.

Then

$$-\frac{1}{2} \leq \frac{x}{x^2 + 1} \leq \frac{1}{2}$$

The second method

$$y = \frac{x}{x^2 + 1},$$

$yx^2 - x + y = 0$ is a quadratic equation in terms of $x$,

$$D = 1 - 4y^2 \geq 0,$$

Hence $y^{2} \leq \frac{1}{4}$ , that is, $-\frac{1}{2} \leq y \leq \frac{1}{2}$ .

The third method.

1) We now find the asymptotic value of the function for $x \to \pm \infty$

$$\lim _ {x \rightarrow + \infty} f (x) = \lim _ {x \rightarrow + \infty} \frac {x}{x ^ {2} + 1} = + 0$$ $$\lim _ {x \rightarrow - \infty} f (x) = \lim _ {x \rightarrow - \infty} \frac {x}{x ^ {2} + 1} = - 0$$

2) Let us find the extremum points

$$\begin{array}{l} f (x) = \frac {x}{x ^ {2} + 1} \rightarrow f ^ {\prime} (x) = \frac {(x) ^ {\prime} \cdot (x ^ {2} + 1) - x \cdot (x ^ {2} + 1) ^ {\prime}}{(x ^ {2} + 1) ^ {2}} = \frac {1 \cdot (x ^ {2} + 1) - x \cdot (2 x)}{(x ^ {2} + 1) ^ {2}} = \\ = \frac {x ^ {2} + 1 - 2 x ^ {2}}{(x ^ {2} + 1) ^ {2}} = \frac {1 - x ^ {2}}{(1 + x ^ {2}) ^ {2}} = \frac {(1 - x) (1 + x)}{(1 + x ^ {2}) ^ {2}} \rightarrow \boxed {f ^ {\prime} (x) = \frac {(1 - x) (1 + x)}{(1 + x ^ {2}) ^ {2}}} \\ \end{array}$$ $$f ^ {\prime} (x) = 0 \rightarrow \frac {(1 - x) (1 + x)}{(1 + x ^ {2}) ^ {2}} = 0 \rightarrow \left[\begin{array}{l}1 - x = 0\\1 + x = 0\end{array}\right. \rightarrow \boxed {\left[\begin{array}{l}x = 1\\x = - 1\end{array}\right]}$$$$\min _ {x \in \mathbb {R}} f (x) = f (- 1) = \frac {- 1}{(- 1) ^ {2} + 1} = - \frac {1}{2}$$ $$\max _ {x \in \mathbb {R}} f (x) = f (1) = \frac {1}{1 ^ {2} + 1} = - \frac {1}{2}$$

3) Let's sketch this function using the received data

As we see from the sketch, the local extrema are global maxima and minima/ Then,

$$y = f (x) = \frac {x}{x ^ {2} + 1} \rightarrow R a n g e: y \in \left[ - \frac {1}{2}; \frac {1}{2} \right]$$

ANSWER:

$$y = f (x), f (x) = \frac {x}{x ^ {2} + 1} \rightarrow D o m a i n: x \in \mathbb {R}$$ $$y = f (x), f (x) = \frac {x}{x ^ {2} + 1} \rightarrow R a n g e: y \in \left[ - \frac {1}{2}; \frac {1}{2} \right].$$

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