Use the IVT to prove that one solution to {3} +x-5=0" alt="x^{3} +x-5=0" align="absmiddle" class="latex-formula"> is between x=1 and x=2. Please show work and if your answer makes no sense I will report. Thanks
1 answer:
Let <em>f(x)</em> = <em>x</em>³ + <em>x</em> - 5. <em>f(x)</em> is a polynomial so it's continuous everywhere on its domain (all real numbers). Since
<em>f</em> (1) = 1³ + 1 - 5 = -3 < 0
and
<em>f</em> (2) = 2³ + 2 - 5 = 5 > 0
it follows by the intermediate value theorem that there at least one number <em>x</em> = <em>c</em> between 1 and 2 for which <em>f(c)</em> = 0.
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