<s>[INST] Find the Range of the function $f(x) = |x-6|+x^2-1$ 
find the Range of $f(x) = |x-6|+x^2-1$

$$ f(x) = |x-6|+x^2-1 =\left\{ 
\begin{array}{c}
x^2+x-7,& x>0 .....(b) \\ 
5,& x=0 .....(a) \\ 
x^2-x+5,& x<0 ......(c)
\end{array}
\right. 
$$
from eq (b) i got $$f(x)= \left(x+\frac12\right)^2-\frac{29}4 \ge-\frac{29}4$$
and from eq (c) i got $$f(x)= \left(x-\frac12\right)^2+\frac{19}4 \ge\frac{19}4$$
and eq(b) tells me that it also passes through 5 and so generalize all this and found its range is $\left[-\frac{29}4 , \infty\right)$
but the graph says its range is $(5, \infty)$
 [/INST] No derivatives are necessary. For $x\ge6$ the function is
$$
f(x)=x^2+x-7
$$
The graph is an arc of a parabola with its axis at $x=-1/2$, so in this interval the function is increasing, with its minimum at $6$: $f(6)=35$.
For $x<6$ the function is
$$
f(x)=x^2-x+5
$$
whose graph is an arc of a parabola with its axis at $x=1/2$. Since $1/2<6$ the minimum of $f$ is reached at $1/2$: $f(1/2)=19/4$.
Thus the range is $(19/4,\infty)$ because, clearly, $\lim_{x\to\infty}f(x)=\infty$.
</s>