3

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 $(5, \infty)$

zoli
  • 20,452
anni
  • 349
  • When you have something like $|x-k|$ in the equation, try to expand it for the two cases of $x \geq k$ and $x \lt k$ instead of expanding around $x \geq 0$ and $x \lt 0$. – wadkar May 13 '16 at 04:26

3 Answers3

3

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, $\lim_{x\to\infty}f(x)=\infty$.

egreg
  • 238,574
2

$$f'(x)=\frac{x-6}{|x-6|}+2x$$

$f'(x)=0\iff x=\frac{1}{2}$ and $f'(x)<0$ if $x<\frac{1}{2}$ and $f'(x)>0$ if $x>\frac{1}{2}$, therefore the range is $[f(\frac{1}{2}),+\infty [$.

Surb
  • 55,662
1

You should have $x-6<0$, $x-6=0$ and $x-6>0$ respectively. Always look to the entire expression within the absolute value.

Oh, and another thing: While finding such minimum, you need to check whether it is in the domain. For example, you have $x>0$ (should be $x>6$) for (b), but the minimum is given at $x=-\frac{1}{2}$.

wythagoras
  • 25,026