How do you solve $(x - 1) (3 x - 4) \geq 0$ $(x-1)(3x-4)>=0$ using a sign chart?

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How do you solve $(x - 1) (3 x - 4) \geq 0$ $(x-1)(3x-4)>=0$ using a sign chart?

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Answer 1 · 2016-12-06T08:32:52

The answer is $x \in] - \infty, 1] \cup [\frac{4}{3}, + \infty [$ $x in ] -oo,1 ] uu [ 4/3, +oo[$

Explanation:

Let $f (x) = (x - 1) (3 x - 4)$ $f(x)=(x-1)(3x-4)$

Let's do the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $1$ $1$ $a a a a$ $color(white)(aaaa)$ $\frac{4}{3}$ $4/3$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x - 1$ $x-1$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $3 x - 4$ $3x-4$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

So, $f (x) \geq 0$ $f(x)>=0$ for $x \in] - \infty, 1] \cup [\frac{4}{3}, + \infty [$ $x in ] -oo,1 ] uu [ 4/3, +oo[$

graph{y=(x-1)(3x-4) [-1.863, 3.006, -1.232, 1.2]}

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How do you solve $(x - 1) (3 x - 4) \geq 0$ $(x-1)(3x-4)>=0$ using a sign chart?

1 Answer

Explanation:

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