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Question: How do you find the critical points of a rational function?...

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

Explanation

Solution

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.

Complete step-by-step answer:
The critical points of a function f(x)  f(x)\; are those where the following conditions apply:
A) The function exists.
B) The derivative of the function f′(x)  f'(x)\; is either equal to 00 or does not exist.
As an example, with a polynomial function, suppose we take the function
f(x)=x2+5x−7f(x) = {x^2} + 5x - 7.
The derivative of this function is, f′(x)=2x+5f'(x) = 2x + 5.
For our first type of critical point, those where the derivative is equal to zero, we simply set the derivative equal to 00.
Therefore,
f’(x)=0f’(x) = 0
⇒2x+5=0\Rightarrow 2x + 5 = 0
⇒2x=−5\Rightarrow 2x = - 5
Now, divide both sides by 22.
⇒x=−2.5\Rightarrow x = - 2.5
Doing this, we can find that the only point where the derivative is 00 at x=−2.5x = - 2.5.
Now, we will find the value of f(x) at x=−2.5f\left( x \right)\,at\,x = - 2.5
f(x)=(−2.5)2+5(−2.5)−7f(x) = {\left( { - 2.5} \right)^2} + 5\left( { - 2.5} \right) - 7
f(x)=6.25−12.5−7f(x) = 6.25 - 12.5 - 7
f(x)=−6.25−7f(x) = - 6.25 - 7
f(x)=−13.25f(x) = - 13.25.
For our second type of critical point, we 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 (−2.5,−13.25)( - 2.5, - 13.25).
Let’s take another example, we will take the function as f(x)=x23f(x) = {x^{\dfrac{2}{3}}}
On Differentiation, we get
f′(x)=(23)⋅x−13  or  f′(x)=(23)1x13f'(x) = (\dfrac{2}{3}) \cdot {x^{ - \dfrac{1}{3}}}\;or\;f'(x) = \left( {\dfrac{2}{3}} \right)\dfrac{1}{{{x^{\dfrac{1}{3}}}}}.
In this example, there are no real numbers for which f′(x)=0f'(x) = 0, but there is one where f′(x)  f'(x)\; does not exist, namely at x=0x = 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 (0,0).(0,0).

Note: There can be more than one critical point depending upon the equation formed after we find the derivative of the function. It is also possible that there can be no such point. In such cases the function formed is not defined at x=0x = 0.