let ( f(x)=\frac{x}{x + 7} ). find the values of ( x ) where ( f^{prime}(x)=6 ).\ngive exact answers (not…

let ( f(x)=\frac{x}{x + 7} ). find the values of ( x ) where ( f^{prime}(x)=6 ).\ngive exact answers (not decimal approximations).\nthe greater solution is ( x=)\nthe lesser solution is ( x=)\nquestion help: video message instructor\nsubmit question jump to answer

let ( f(x)=\frac{x}{x + 7} ). find the values of ( x ) where ( f^{prime}(x)=6 ).\ngive exact answers (not decimal approximations).\nthe greater solution is ( x=)\nthe lesser solution is ( x=)\nquestion help: video message instructor\nsubmit question jump to answer

Answer

Explanation:

Step1: Differentiate ( f(x) ) using quotient rule

The quotient rule states that if ( f(x)=\frac{u}{v} ), then ( f^{\prime}(x)=\frac{u^{\prime}v - uv^{\prime}}{v^{2}} ). Here ( u = x), (u^{\prime}=1), (v=x + 7), (v^{\prime}=1). [ \begin{align*} f^{\prime}(x)&=\frac{1\times(x + 7)-x\times1}{(x + 7)^{2}}\ &=\frac{x+7 - x}{(x + 7)^{2}}\ &=\frac{7}{(x + 7)^{2}} \end{align*} ]

Step2: Set ( f^{\prime}(x)=6 ) and solve for ( x )

Set (\frac{7}{(x + 7)^{2}}=6). Cross - multiply to get (7 = 6(x + 7)^{2}). Then ((x + 7)^{2}=\frac{7}{6}). Take square roots: (x+7=\pm\sqrt{\frac{7}{6}}=\pm\frac{\sqrt{42}}{6}). Solve for (x): (x=-7\pm\frac{\sqrt{42}}{6}).

Answer:

The greater solution is (x=-7+\frac{\sqrt{42}}{6}). The lesser solution is (x=-7-\frac{\sqrt{42}}{6}).