What are the critical points of f(x) = -(2x - 2.7)/ (2 x -7.29 )^2f(x)=−2x−2.7(2x−7.29)2?
1 Answer
Explanation:
We're asked to find the critical points of the function
f(x) = -(2x-2.7)/((2x-7.29)^2f(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)]ddx[−2x−2.7(2x−7.29)2]
Which is the same as
d/(dx) [(2.7-2x)/((2x-7.29)^2)]ddx[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)ddx[uv]=vdudx+udvdx
where
-
u = 2.7-2xu=2.7−2x -
v = 1/((2x-7.29)^2)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)]=ddx[2.7−2x](2x−7.29)2+(2.7−2x)ddx[1(2x−7.29)2]
= (-2)/((2x-7.29)^2) + (2.7-2x) d/(dx) [1/((2x-7.29)^2)]=−2(2x−7.29)2+(2.7−2x)ddx[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)ddx[1(2x−7.29)2]=ddu[1u2]dudx
where
-
u = 2x-7.29u=2x−7.29 -
d/(du) [1/(u^2)] = (-2)/(u^3)ddu[1u2]=−2u3 :
= (-2)/((2x-7.29)^2) + (2.7-2x)((-2d/(dx)[2x-7.29])/((2x-7.29)^3))=−2(2x−7.29)2+(2.7−2x)(−2ddx[2x−7.29](2x−7.29)3)
= color(blue)((-2)/((2x-7.29)^2) + (2.7-2x)((-4)/((2x-7.29)^3))=−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)=−2(2x−7.29)2−4(2.7−2x)(2x−7.29)3
Getting a common denominator of
= (-2(2x-7.29))/((2x-7.29)^3) - (4(2.7-2x))/((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)=−4x+14.58+8x−10.8(2x−7.29)3
= color(blue)((4x + 3.78)/((2x-7.29)^3)=4x+3.78(2x−7.29)3
The critical points can be found by setting the numerator equal to
4x+3.78 = 04x+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).
