Question

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

Explanation:

Step1: Differentiate \( f(x) \) using the 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, \( u = x \), \( u^{\prime}=1 \), \( v=x + 7 \), \( v^{\prime}=1 \).
So \( f^{\prime}(x)=\frac{1\cdot(x + 7)-x\cdot1}{(x + 7)^{2}}=\frac{x + 7-x}{(x + 7)^{2}}=\frac{7}{(x + 7)^{2}} \).

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 \( x=-7-\frac{\sqrt{42}}{6} \).