Question

The slope of the tangent line to the function f(x) = 8/(x-3) at x = –5 is–1.-1/81/88.

Answer 1

The slope is -1/8

EXPLANATION :

From the problem, we have the function :

`f(x)=(8)/(x-3)`

The slope of the tangent line at x = -5 is the value of the first derivative at x = -5

Differentiate the function :

`\begin{gathered} f(x)=(8)/(x-3) \\ \\ \text{ can be written as :} \\ \\ f(x)=8(x-3)^(-1) \\ \\ \text{ using the general rule for differentiation :} \\ \\ f(x)=ax^n \\ f^(\prime)(x)=n(ax^(n-1)) \\ \\ \text{ Then :} \\ \\ f^(\prime)(x)=-1[8(x-3)^(-1-1)] \\ f^(\prime)(x)=-8(x-3)^(-2) \\ \\ \text{ Write as a fraction :} \\ f^(\prime)(x)=-(8)/((x-3)^2) \\ \\ \text{ Then substitute x = -5} \\ \\ f^(\prime)(-5)=-(8)/((-5-3)^2)=-(8)/((-8)^2) \\ \\ f^(\prime)(-5)=-(8)/(64)=-(1)/(8) \end{gathered}`