Problems from Graphic representation of a function

Development:

1) The function does not contain any problems in its definition, so the domain is the whole real line and the image too. We will find the table giving $$3$$ points and evaluating them.

2) The function does not contain any problems in its definition, so the domain is the whole real line. On the other hand, since it appears $$x^2$$ we will only obtain negative values when $$x^2< 2$$. The image will be the interval $$[-2,\infty)$$. We will do the table in the same way.

3) As we have a division, it is possible that we have some problems when we find the value $$0$$ in the denominator. Therefore: $$x+1=0\Leftrightarrow x=-1$$, and we see that the domain is $$\mathbb{R}\setminus\{-1\}$$. The image will be all the real values except zero, since we will never be able to reach it. We will do the table with enough points because the function is curved and can turn out to be difficult to plot.

Solution:

1) $$\text{Dom}(f)=\mathbb{R} \quad$$ and $$\quad \text{Im}(f)=\mathbb{R}$$.

$$x$$	$$f(x) = -x+2$$
$$-1$$	$$3$$
$$0$$	$$2$$
$$1$$	$$1$$

2) $$\text{Dom}(f)=\mathbb{R} \quad$$ and $$\quad \text{Im}(f)=[-2,\infty)$$.

$$x$$	$$f(x)=x^2-2$$
$$-2$$	$$2$$
$$-1$$	$$-1$$
$$0$$	$$-2$$
$$1$$	$$-1$$
$$2$$	$$2$$
$$3$$	$$7$$

3) $$\text{Dom}(f)=\mathbb{R}\setminus\{-1\} \quad$$ and $$\quad \text{Im}(f)=\mathbb{R}\setminus\{-1\}$$.

$$x$$	$$f(x)=\dfrac{1}{x+1}$$
$$-3$$	$$-0.5$$
$$-2$$	$$-1$$
$$-1.5$$	$$-2$$
$$-1.2$$	$$-5$$
$$-1.1$$	$$-10$$
$$-0.9$$	$$10$$
$$-0.8$$	$$5$$
$$-0.5$$	$$2$$
$$0$$	$$1$$
$$1$$	$$0.5$$

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