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

Aug 5, 2017

$x = - 0.945$

#### 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

$\frac{d}{\mathrm{dx}} \left[- \frac{2 x - 2.7}{{\left(2 x - 7.29\right)}^{2}}\right]$

Which is the same as

$\frac{d}{\mathrm{dx}} \left[\frac{2.7 - 2 x}{{\left(2 x - 7.29\right)}^{2}}\right]$

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

$\frac{d}{\mathrm{dx}} \left[u v\right] = v \frac{\mathrm{du}}{\mathrm{dx}} + u \frac{\mathrm{dv}}{\mathrm{dx}}$

where

• $u = 2.7 - 2 x$

• $v = \frac{1}{{\left(2 x - 7.29\right)}^{2}}$:

$= \frac{\frac{d}{\mathrm{dx}} \left[2.7 - 2 x\right]}{{\left(2 x - 7.29\right)}^{2}} + \left(2.7 - 2 x\right) \frac{d}{\mathrm{dx}} \left[\frac{1}{{\left(2 x - 7.29\right)}^{2}}\right]$

$= \frac{- 2}{{\left(2 x - 7.29\right)}^{2}} + \left(2.7 - 2 x\right) \frac{d}{\mathrm{dx}} \left[\frac{1}{{\left(2 x - 7.29\right)}^{2}}\right]$

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

$\frac{d}{\mathrm{dx}} \left[\frac{1}{{\left(2 x - 7.29\right)}^{2}}\right] = \frac{d}{\mathrm{du}} \left[\frac{1}{{u}^{2}}\right] \frac{\mathrm{du}}{\mathrm{dx}}$

where

• $u = 2 x - 7.29$

• $\frac{d}{\mathrm{du}} \left[\frac{1}{{u}^{2}}\right] = \frac{- 2}{{u}^{3}}$:

$= \frac{- 2}{{\left(2 x - 7.29\right)}^{2}} + \left(2.7 - 2 x\right) \left(\frac{- 2 \frac{d}{\mathrm{dx}} \left[2 x - 7.29\right]}{{\left(2 x - 7.29\right)}^{3}}\right)$

= 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:

$= \frac{- 2}{{\left(2 x - 7.29\right)}^{2}} - \frac{4 \left(2.7 - 2 x\right)}{{\left(2 x - 7.29\right)}^{3}}$

Getting a common denominator of ${\left(2 x - 7.29\right)}^{3}$:

$= \frac{- 2 \left(2 x - 7.29\right)}{{\left(2 x - 7.29\right)}^{3}} - \frac{4 \left(2.7 - 2 x\right)}{{\left(2 x - 7.29\right)}^{3}}$

$= \frac{- 4 x + 14.58 + 8 x - 10.8}{{\left(2 x - 7.29\right)}^{3}}$

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

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

$4 x + 3.78 = 0$

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).