Problems from Image of a function

Development:

1) If we compute the vertex of the parable:

v:$(-{\displaystyle \frac{b}{2a}},-{\displaystyle \frac{b^2-4ac}{4a}})=(0,-2)$

and since $a=1>0$, the parabola is convex (or concave) and therefore we have $Im(f)=[-2,+\mathrm{\infty })$

2) We know that square roots have the following image: $Im(f)=[0,+\mathrm{\infty })$ (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, $Im(f)=\mathbb{R}-\{0\}$

Solution:

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

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

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

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

3) $f(x)={\displaystyle \frac{1}{x+1}}$

$Im(f)=\mathbb{R}-\{0\}$

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