What are the critical points of #f(x) = -(2x - 2.7)/ (2 x -7.29 )^2#?

1 Answer
Aug 5, 2017

Answer:

#x = -0.945#

Explanation:

We're asked to find the critical points of the function

#f(x) = -(2x-2.7)/((2x-7.29)^2#

A critical point of a function is a point where the first derivative is equal to zero or is undefined (but the function must be defined at that point).

With that being said, let's find the derivative

#d/(dx) [-(2x-2.7)/((2x-7.29)^2)]#

Which is the same as

#d/(dx) [(2.7-2x)/((2x-7.29)^2)]#

First, let's use the product rule, which states

#d/(dx) [uv] = v(du)/(dx) + u(dv)/(dx)#

where

  • #u = 2.7-2x#

  • #v = 1/((2x-7.29)^2)#:

#= (d/(dx) [2.7-2x])/((2x-7.29)^2) + (2.7-2x) d/(dx) [1/((2x-7.29)^2)]#

#= (-2)/((2x-7.29)^2) + (2.7-2x) d/(dx) [1/((2x-7.29)^2)]#

Now, we'll use the chain rule on the second term**, which here is

#d/(dx)[1/((2x-7.29)^2)] = d/(du) [1/(u^2)] (du)/(dx)#

where

  • #u = 2x-7.29#

  • #d/(du) [1/(u^2)] = (-2)/(u^3)#:

#= (-2)/((2x-7.29)^2) + (2.7-2x)((-2d/(dx)[2x-7.29])/((2x-7.29)^3))#

#= color(blue)((-2)/((2x-7.29)^2) + (2.7-2x)((-4)/((2x-7.29)^3))#

We can continue to simplify this if we'd like to:

#= (-2)/((2x-7.29)^2) - (4(2.7-2x))/((2x-7.29)^3)#

Getting a common denominator of #(2x-7.29)^3#:

#= (-2(2x-7.29))/((2x-7.29)^3) - (4(2.7-2x))/((2x-7.29)^3)#

#= (-4x + 14.58 + 8x - 10.8)/((2x-7.29)^3)#

#= color(blue)((4x + 3.78)/((2x-7.29)^3)#

The critical points can be found by setting the numerator equal to #0#:

#4x+3.78 = 0#

#color(red)(ulbar(|stackrel(" ")(" "x = -0.945" ")|)#

We can't set the denominator equal to zero, because the function itself is undefined at that point (so it is not a critical point).