14_Cal_Solution of Calculus_6e

# 14_Cal_Solution of Calculus_6e - -4-2246-4-224C01S02.035:To...

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Unformatted text preview: -4-2246-4-224C01S02.035:To graphf(x) =√x2−9, note that there is no graph for−3&lt; x &lt;3, thatf(±3) = 0, andthatf(x)&gt;0 forx &lt;−3 and forx &gt;3. Ifxis large positive, then√x2−9≈√x2=x, so the graph offhasx-intercept (3,0) and rises asxincreases, nearly coinciding with the graph ofy=xforxlarge positive.The casex &lt;−3 is trickier. In this case, ifxis a large negative number, thenf(x) =√x2−9≈√x2=−x(Note the minus sign!). So forx5−3, the graph offhasx-intercept (−3,0) and, forxlarge negative,almost coincides with the graph ofy=−x. Later we will see that the graph offbecomes arbitrarily steepasxgets closer and closer to±3.C01S02.036:Asxincreases without bound—either positively or negatively—f(x) gets arbitrarily closeto zero. Moreover, ifxis large positive thenf(x) is negative and close to zero, so the graph offlies justbelow thex-axis for suchx. Similarly, the graph offlies just above thex-axis forxlarge negative. Ifxis slightly less than 1 but very close to 1, then...
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## This note was uploaded on 09/14/2011 for the course MATH 101 taught by Professor Cheng during the Spring '11 term at Ecole Hôtelière de Lausanne.

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