How do you find the slope of a tangent line to the graph of the function f(x)=8−3x−1 at x=-3?

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fonne0kgq · Accepted Answer

By definition, given a function f(x), is derivative f'(x) associates with every point x the slope of the tangent line to f in x.
In other words, the answer is f'(-3). Let's compute it.
Since

\frac{d}{d x} \sqrt{f (x)} = \frac{f^{'} (x)}{2 \sqrt{f (x)}}

, we have

\frac{d}{d x} 8 \sqrt{- 3 x - 1} = 8 \frac{- 3}{2 \sqrt{- 3 x - 1}} = \frac{- 12}{\sqrt{- 3 x - 1}}

Computing the derivative at x=-3, we have

f^{'} (- 3) = \frac{- 12}{\sqrt{- 3 \cdot (- 3) - 1}} = \frac{- 12}{\sqrt{8}}

Simplifying:

- \frac{12}{\sqrt{8}} = - \frac{12}{2 \sqrt{2}} = - \frac{12 \sqrt{2}}{2 \sqrt{2} \sqrt{2}} = - \frac{12 \sqrt{2}}{4} = - 3 \sqrt{2}

How do you find the slope of a

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