Given the functions,
1) $$f(x)=x^2-2$$
2) $$f(x)=\sqrt{x+4}$$
3) $$f(x)=\dfrac{1}{x+1}$$
Determine the image of each of them.
See development and solution
Development:
1) If we compute the vertex of the parable:
v:$$\Big( -\dfrac{b}{2a}, -\dfrac{b^2-4ac}{4a} \Big)=(0,-2) $$
and since $$a = 1> 0$$, the parabola is convex (or concave) and therefore we have $$Im (f) = [-2, +\infty)$$
2) We know that square roots have the following image: $$Im (f) = [0, +\infty)$$ (since we take the positive solution of the square root).
3) We can see then that we can obtain any real number except zero. Therefore, $$Im (f) = \mathbb{R} - \lbrace0\rbrace$$
Solution:
1) $$f(x)=x^2-2$$
$$Im (f) = [-2, +\infty)$$
2) $$f(x)=\sqrt{x+4}$$
$$Im (f) = [0, +\infty)$$
3) $$f(x)=\dfrac{1}{x+1}$$
$$Im (f) = \mathbb{R} - \lbrace0\rbrace$$