Use an indirect proof to show that if x 3 + x − 1 &gt; 10 then x &gt; 1.

Question

Use an indirect proof to show that if       x    3    +  x  −  1  &amp;gt;  10 then   x  &amp;gt;  1.

h2a2l1i2morz · Accepted Answer

If   x  ≤  1 then either   x  ≤  0 or   0  &amp;lt;  x  ≤  1. If   x  ≤  0 then obviously       x    3    ≤  0  ≤  1. If   0  &amp;lt;  x  ≤  1 then multiply the inequality three times with itself (you can do that since both   x and   1 are positive numbers) and you get that       x    3    ≤  1.Thus in any case:   x  ≤  1 implies       x    3    ≤  1. Summation by parts gives      x    3    +  x  −  1  ≤  1Thus      x    3    +  x  −  1  &amp;gt;  10  &amp;gt;  1  ⇒  x  &amp;gt;  1

Mark Rosales · Accepted Answer

Suppose   x  &amp;lt;  1, then   x  −  1  &amp;lt;  0    ⟹    0  &amp;gt;  (  x  −  1  )      [    (    x    +          1      2              )              2              +          3      4        ]    =      x          3        −  1.

Use an indirect proof to show that if x^(3)+x−1>10 then x>1.

Answered question

Answer & Explanation

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