How do you solve $(x + 1) (x - 7) \leq 0$ $(x+1)(x-7)<=0$ using a sign chart?

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How do you solve $(x + 1) (x - 7) \leq 0$ $(x+1)(x-7)<=0$ using a sign chart?

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Answer 1 · 2017-02-19T11:04:09

The solution is $x \in [- 1, 7]$ $x in [-1,7]$

Explanation:

Let $f (x) = (x + 1) (x - 7)$ $f(x)=(x+1)(x-7)$

Now, we build the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a$ $color(white)(aaaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 1$ $-1$ $a a a a$ $color(white)(aaaa)$ $7$ $7$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

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

$a a a a$ $color(white)(aaaa)$ $x - 7$ $x-7$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

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

Therefore,

$f (x) \leq 0$ $f(x)<=0$ when $x \in [- 1, 7]$ $x in [-1,7]$

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How do you solve $(x + 1) (x - 7) \leq 0$ $(x+1)(x-7)<=0$ using a sign chart?

1 Answer

Explanation:

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How do you solve $(x + 1) (x - 7) \leq 0$ $(x+1)(x-7)<=0$ using a sign chart?