What is the domain and range of $f (x) = \sqrt{4 x - x^{2}}$ $f(x)= sqrt(4x-x^2)$ ?

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What is the domain and range of $f (x) = \sqrt{4 x - x^{2}}$ $f(x)= sqrt(4x-x^2)$ ?

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Answer 1 · 2017-07-31T12:29:02

The domain is $x \in [0, 4]$ $x in [0,4]$
The range is $f (x) \in [0, 2]$ $f(x) in [0,2]$

Explanation:

For the domain, what's under the square root sign is $\geq 0$ $>=0$

Therefore,

$4 x - x^{2} \geq 0$ $4x-x^2>=0$

$x (4 - x) \geq 0$ $x(4-x)>=0$

Let $g (x) = \sqrt{x (4 - x)}$ $g(x)=sqrt(x(4-x))$

We can build a sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a a a a$ $color(white)(aaaaaaa)$ $0$ $0$ $a a a a a a$ $color(white)(aaaaaa)$ $4$ $4$ $a a a a a a a$ $color(white)(aaaaaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a a a a$ $color(white)(aaaaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $0$ $0$ $a a$ $color(white)(aa)$ $+$ $+$ $a a a a a a a$ $color(white)(aaaaaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $4 - x$ $4-x$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a$ $color(white)(aa)$ $0$ $0$ $a a a a$ $color(white)(aaaa)$ $-$ $-$

$a a a a$ $color(white)(aaaa)$ $g (x)$ $g(x)$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a$ $color(white)(a)$ $a a a$ $color(white)(aaa)$ $0$ $0$ $a a$ $color(white)(aa)$ $+$ $+$ $a a$ $color(white)(aa)$ $0$ $0$ $a a a a$ $color(white)(aaaa)$ $-$ $-$

Therefore

$g (x) \geq 0$ $g(x)>=0$ when $x \in [0, 4]$ $x in [0,4]$

Let,

$y = \sqrt{4 x - x^{2}}$ $y=sqrt(4x-x^2)$

hen,

$y^{2} = 4 x - x^{2}$ $y^2=4x-x^2$

$x^{2} - 4 x + y^{2} = 0$ $x^2-4x+y^2=0$

The solutions this quadratic equation is when the discriminant $Delta>=0$

So,

$Delta=(-4)^2-4*1*y^2$

$16-4y^2>=0$

$4(4-y^2)>=0$

$4(2+y)(2-y)>=0$

Let $h(y)=(2+y)(2-y)$

We build the sign chart

$color(white)(aaaa)$ $y$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaaa)$ $-2$ $color(white)(aaaa)$ #color(white)(aaaaaa)2# $color(white)(aaaaaa)$ $+oo$

$color(white)(aaaa)$ $2+y$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $2-y$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $-$

$color(white)(aaaa)$ $h(y)$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $-$

Therefore,

$h(y)>=0$ , when $y in [-2,2]$

This is not possible for the whole interval, so the range is $y in [0,2]$

graph{sqrt(4x-x^2) [-10, 10, -5, 5]}

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What is the domain and range of $f (x) = \sqrt{4 x - x^{2}}$ $f(x)= sqrt(4x-x^2)$ ?

1 Answer

Explanation:

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What is the domain and range of $f (x) = \sqrt{4 x - x^{2}}$ $f(x)= sqrt(4x-x^2)$ ?