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?

Question

Emma Aldridge · Accepted Answer

#-3sqrt(2)#

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)#

Axel Alvarez · Answer

undefined To find the slope of a tangent line to the graph of a function at a specific point, you can use the derivative of the function. In this case, to find the slope of the tangent line to the graph of f(x) = 8√(-3x - 1) at x = -3, you need to follow these steps:

1. Find the derivative of the function f(x) with respect to x.
2. Substitute the value x = -3 into the derivative to find the slope at that point.

Let's calculate the derivative of f(x) = 8√(-3x - 1):

f'(x) = d/dx [8√(-3x - 1)]

Using the chain rule, we have:

f'(x) = 8 * (1/2) * (-3x - 1)^(-1/2) * (-3)

Now, substitute x = -3 into the derivative:

f'(-3) = 8 * (1/2) * (-3(-3) - 1)^(-1/2) * (-3)

Simplifying further:

f'(-3) = 8 * (1/2) * (-8)^(-1/2) * (-3)

Finally, calculate the value:

f'(-3) = -12/√8

Therefore, the slope of the tangent line to the graph of f(x) = 8√(-3x - 1) at x = -3 is -12/√8.