What are the critical points of #f(x) = (2x  2.7)/ (2 x 7.29 )^2#?
1 Answer
Answer:
Explanation:
We're asked to find the critical points of the function
#f(x) = (2x2.7)/((2x7.29)^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) [(2x2.7)/((2x7.29)^2)]#
Which is the same as
#d/(dx) [(2.72x)/((2x7.29)^2)]#
First, let's use the product rule, which states
#d/(dx) [uv] = v(du)/(dx) + u(dv)/(dx)#
where

#u = 2.72x# 
#v = 1/((2x7.29)^2)# :
#= (d/(dx) [2.72x])/((2x7.29)^2) + (2.72x) d/(dx) [1/((2x7.29)^2)]#
#= (2)/((2x7.29)^2) + (2.72x) d/(dx) [1/((2x7.29)^2)]#
Now, we'll use the chain rule on the second term**, which here is
#d/(dx)[1/((2x7.29)^2)] = d/(du) [1/(u^2)] (du)/(dx)#
where

#u = 2x7.29# 
#d/(du) [1/(u^2)] = (2)/(u^3)# :
#= (2)/((2x7.29)^2) + (2.72x)((2d/(dx)[2x7.29])/((2x7.29)^3))#
#= color(blue)((2)/((2x7.29)^2) + (2.72x)((4)/((2x7.29)^3))#
We can continue to simplify this if we'd like to:
#= (2)/((2x7.29)^2)  (4(2.72x))/((2x7.29)^3)#
Getting a common denominator of
#= (2(2x7.29))/((2x7.29)^3)  (4(2.72x))/((2x7.29)^3)#
#= (4x + 14.58 + 8x  10.8)/((2x7.29)^3)#
#= color(blue)((4x + 3.78)/((2x7.29)^3)#
The critical points can be found by setting the numerator equal to
#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).