Image of a function

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.

  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 $Im (f) = [-2, +\infty)$

  1. We know that square roots have the following image: $Im (f) = [0, +\infty)$ (since we take the positive solution of the square root).

  2. We can see then that we can obtain any real number except zero. Therefore, $Im (f) = \mathbb{R} - \lbrace0\rbrace$

  1. $f(x)=x^2-2$

$$Im (f) = [-2, +\infty)$$

  1. $f(x)=\sqrt{x+4}$

$$Im (f) = [0, +\infty)$$

  1. $f(x)=\dfrac{1}{x+1}$

$$Im (f) = \mathbb{R} - \lbrace0\rbrace$$

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