How do you find the domain of $f (x) = \sqrt{x - (3 x^{2})}$ $f(x) = sqrt ( x- (3x^2))$ ?

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How do you find the domain of $f (x) = \sqrt{x - (3 x^{2})}$ $f(x) = sqrt ( x- (3x^2))$ ?

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Answer 1 · 2018-03-18T11:07:33

The domain of $f (x)$ $f(x)$ is $x \in [0, \frac{1}{3}]$ $x in [0,1/3]$

Explanation:

What's under the $\sqrt{}$ $sqrt()$ sign is $\geq 0$ $>=0$

Here,

$f (x) = \sqrt{x - 3 x^{2}}$ $f(x)=sqrt(x-3x^2)$

Therefore,

$x - 3 x^{2} \geq 0$ $x-3x^2>=0$

$x (1 - 3 x) \geq 0$ $x(1-3x)>=0$

Let $g (x) = x (1 - 3 x)$ $g(x)=x(1-3x)$

Build the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a a a$ $color(white)(aaaaaa)$ $0$ $0$ $a a a a a a a$ $color(white)(aaaaaaa)$ $\frac{1}{3}$ $1/3$ $a a a a a a$ $color(white)(aaaaaa)$ $+ \infty$ $+oo$

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

$a a a a$ $color(white)(aaaa)$ $1 - 3 x$ $1-3x$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ #color(white)(aaaa)+# $a a$ $color(white)(aa)$ $0$ $0$ $a a a a$ $color(white)(aaaa)$ $-$ $-$

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

Therefore,

$g (x) \leq 0$ $g(x)<=0$ , $\Rightarrow$ $=>$ , $x \in [0, \frac{1}{3}]$ $x in [0,1/3]$

graph{sqrt(x-3x^2) [-0.589, 1.096, -0.307, 0.536]}

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How do you find the domain of $f (x) = \sqrt{x - (3 x^{2})}$ $f(x) = sqrt ( x- (3x^2))$ ?

1 Answer

Explanation:

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How do you find the domain of f(x) = sqrt ( x- (3x^2))f(x)=√x−(3x2)f(x) = sqrt ( x- (3x^2))?

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How do you find the domain of $f (x) = \sqrt{x - (3 x^{2})}$ $f(x) = sqrt ( x- (3x^2))$ ?