# How do you find the critical points of a rational function?

Aug 16, 2014

To find the critical points of a function, first ensure that the function is differentiable, and then take the derivative. Next, find all values of the function's independent variable for which the derivative is equal to 0, along with those for which the derivative does not exist. These are our critical points.

The critical points of a function $f \left(x\right)$ are those where the following conditions apply:

A) The function exists.

B) The derivative of the function $f ' \left(x\right)$ is either equal to 0 or does not exist.

As an example with a polynomial function, suppose I take the function $f \left(x\right) = {x}^{2} + 5 x - 7$ The derivative of this function, according to the power rule, is the function $f ' \left(x\right) = 2 \cdot x + 5$.

For our first type of critical point, those where the derivative is equal to zero, I simply set the derivative equal to 0. Doing this, I find that the only point where the derivative is 0 is at $x = - 2.5$, at which value $f \left(x\right) = - 13.25$.

For our second type of critical point, I look to see if there are any values of $x$ for which my derivative does not exist. I see there are none, so I am confident in stating that the only critical point on my function occurs at $\left(- 2.5 , - 13.25\right)$

For a slightly more tricky example, we will take the function $f \left(x\right) = {x}^{\frac{2}{3}}$ Differentiation yields $f ' \left(x\right) = \left(\frac{2}{3}\right) \cdot {x}^{- \frac{1}{3}}$ or $f ' \left(x\right) = \frac{2}{3 {x}^{\frac{1}{3}}}$. In this example, there are no real numbers for which $f ' \left(x\right) = 0$, but there is one where $f ' \left(x\right)$ does not exist, namely at $x = 0$. The original function, however, does exist at this point, thus satisfying condition A from the summary. Therefore, this function possesses a critical point at $\left(0 , 0\right)$.