Thread: Proving statement for calculus, one step away. Help!

1. Proving statement for calculus, one step away. Help!

Prove (x+1)^2 < or equal to 1 implies x< or equal to 0

Taking left hand side:

(x+1)^2 is less than or equal to 1

x^2 +2x < or equal to 0

So x(x+2) < or equal to 0.

If x =0, statemtn is true as 0.2 = 0

Is that accurate working out. How do I do the last step?!

2. Hello, chr91!

You're off to a good start . . .

$\text{Prove }(x+1)^2 \:\le\:1\,\text{ implies }\,x \:\le\:0$

$\text{Taking left hand side:}$

. . $(x+1)^2 \:\le\:1$

. . $x^2 +2x \:\le\:0$

. . $\text{So: }\:x(x+2) \:\le\:0$

The expression equals 0 when $x \,=\,0,\:\text{-}2$

The number line is divided into three intervals:

. . $\begin{array}{ccccc}\underbrace{-----} & * & \underbrace{----} & * & \underbrace{-----} \\ & \text{-}2 & & 0 & \end{array}$

Test a value in each interval; see if it satisfies the inequality.

$\begin{array}{ccccccccc}
\text{In }(\text{-}\infty,\text{-}2), & \text{test }x = \text{-}3\!: & (\text{-}3)(\text{-}3+2) &=& \text{-}3(\text{-}1) &=& +3 & \text{no} \\

\text{In }(\text{-}2,\,0), & \text{test }x = \text{-}1\!: & (\text{-}1)(\text{-}1+2) & =& (\text{-}1)(1) &=& -1 & \text{yes} \\

\text{In }(0,\infty). & \text{test }x = 1\!: & (1)(1+2) &=& (1)(3) &=& +3 & \text{no} \end{array}$

The inequality is satisfied on the interval $(-2,\,0)$

Therefore: . $-2 \:\leq\: x \:\leq 0$