# Thread: Finding range of fuctions.

1. ## Finding range of fuctions.

hello.. Could someone please show me how to find the range of these equations. Thanks ~~

Find domain for the following relation:
a) (x + 2)^2 + (y-3)^2 = 49

Find the range of the following functions:
a) m(x) = (x^2 + 2x +5)
b) (x – 2)^2 + (y+1)^2 = 64
c) h(x) = √(x+4) + √(5-x)

Just some homework qus i m stuck with and cant work out... Thanks for your help

2. Originally Posted by tom_asian
hello.. Could someone please show me how to find the range of these equations. Thanks ~~

Find domain for the following relation:
a) (x + 2)^2 + (y-3)^2 = 49

Find the range of the following functions:
a) m(x) = (x^2 + 2x +5)
b) (x – 2)^2 + (y+1)^2 = 64
c) h(x) = √(x+4) + √(5-x)

Just some homework qus i m stuck with and cant work out... Thanks for your help
The simplest way to find the range is to graph the relation/function.
a) $(x + 2)^2 + (y - 3)^2 = 49$
This is a circle of radius 7, with center (-2, 3), so the y value is between 3 - 7 = -4 and 3 + 7 = 10. Thus the range is $[-4, 10 ]$.

a) $m(x) = \sqrt{x^2 + 2x +5}$
This can be modified to
$m(x) = \sqrt{(x + 1)^2 + 4}$
Thus the minimum value of the quadratic function is 4, so the minimum m value is 2. There is no upper limit on the radicand, so the upper "value" of m is $\infty$. Thus the range is $[2, \infty )$.

b) $(x - 2)^2 + (y + 1)^2 = 64$
This is a circle of radius 8, with center (2, -1). You do this one.

c) $h(x) = \sqrt{x + 4} + \sqrt{5 - x}$
This is a bit trickier to see what to do, so I graphed it. (See below.)
The minimum value of the range is apparently 3. The peak of the function is at x = 1/2 (we can prove this with Calculus, but you can also see it on the graph), so the maximum y value is $h \left ( \frac{1}{2} \right ) = \sqrt{\frac{1}{2} + 4} + \sqrt{5 - \frac{1}{2}} = 3 \sqrt{2}$
So the range is $[ 3, 3\sqrt{2} ]$

-Dan

3. Originally Posted by tom_asian
c) h(x) = √(x+4) + √(5-x)
Here is a purely algebraic way of doing this one.

The domain of this function is $-4\leq x\leq 5$.

Consider the equation for $y$:
$y=\sqrt{x+4}+\sqrt{5-x}$
The question is for what $y$ does this have at least one real solution for $x\mbox{ on }[-4,5]$.

Thus,
$y^2 = x+4 + 5 - x + 2\sqrt{(x+4)(5-x)}$
$y^2 - 9 = 2\sqrt{20+x-x^2}$
$(y^2-9)^2 = 80+4x-4x^2$
$-4x^2+4x+[80-(y^2-9)^2]=0$
In order to have a real solution we need,
$16 + 16[80-(y^2-9)^2]\geq 0$
$81 - (y^2-9)^2 \geq 0$
Thus,
$(y^2-9)^2 \leq 81$
With the solution,
$x=\frac{-4\pm \sqrt{16+16[80-(y^2-9)^2]}}{-8}$
$x=\frac{1\pm \sqrt{81-(y^2-9)^2}}{2}$
But we want,
$-4\leq 1-\sqrt{81-(y^2-9)^2} \mbox{ and }1+\sqrt{81-(y^2-9)^2} \leq 5$
If you solve these inequalities system (which is not that hard) you get what topsquark got.