Finding p so that[-1,-13]∉range&nbsp;fwhere(p−x2+1)f(x)=x−1,x≠±p+1

Question

Finding p so that[-1,-13]∉range&amp;nbsp;fwhere(p−x2+1)f(x)=x−1,x≠±p+1

analiticozuod · Accepted Answer

Step 1Notice that&amp;nbsp;f(x)→0−&amp;nbsp;as&amp;nbsp;x→∞, and&amp;nbsp;f(x)→0+&amp;nbsp;as&amp;nbsp;x→−∞. Also, it is continuous for&amp;nbsp;p≤0&amp;nbsp;so the only way that your condition will be true is if the global minimum value is greater than&amp;nbsp;−13f′(x)=(p−x2−1)−(x−1)(−2x)(p−x−1)2=0⇒x=1±−pNote that if p&amp;gt;0 then there would be no local maxima/minima suggesting the existence of a vertical asymptote, and hence our condition would not be satisfied.So,&amp;nbsp;p≤0. Now it is easy to check that f(x) attains it’s minimum value at&amp;nbsp;x=1+−p&amp;nbsp;and we wantf(1+−p)&amp;gt;−13−p2p−2−p&amp;gt;−133−p&amp;lt;2−p−2p−p&amp;lt;4p2⇒p&amp;lt;-14and we are done.

Finding p so that \(\displaystyle{\left[-{1},-{\frac{{{1}}}{{{3}}}}\right]}\cancel{\in}\text{range }\ {f}\) where

Answered question

Answer & Explanation

New Questions in Algebra I