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Math Help - Proving statement for calculus, one step away. Help!

  1. #1
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    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?!
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  2. #2
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    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}<br />
\text{In }(\text{-}\infty,\text{-}2), & \text{test }x = \text{-}3\!: & (\text{-}3)(\text{-}3+2) &=& \text{-}3(\text{-}1) &=& +3 & \text{no} \\<br /> <br />
\text{In }(\text{-}2,\,0), & \text{test }x = \text{-}1\!: & (\text{-}1)(\text{-}1+2) & =& (\text{-}1)(1) &=& -1 & \text{yes} \\<br /> <br />
\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
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