Question

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

Answer 1

`-3sqrt(2)`

Explanation:

By definition, given a function `f(x)`, its 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 `d/dx sqrt(f(x))=(f'(x))/(2sqrt(f(x)))`, we have

`d/dx 8sqrt(-3x-1) = 8 (-3)/(2sqrt(-3x-1))=(-12)/(sqrt(-3x-1))`

Computing the derivative at `x=-3`, we have

`f'(-3) = (-12)/(sqrt(-3*(-3)-1)) = (-12)/sqrt(8)`

Simplifying:

`-12/sqrt(8)=-12/(2sqrt(2))=-(12sqrt(2))/(2sqrt(2)sqrt(2))=-(12sqrt(2))/4=-3sqrt(2)`