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

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

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ANSWER :

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}

User Assafmo
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