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Critical Point
The concept of critical point is very important in Calculus as it is used widely in solving optimization problems. The graph of a function has either a horizontal tangent or a vertical tangent at the critical point. Based upon this we will derive a few more facts about critical points.
Let us learn more about critical points along with its definition and how to find it from a function and from a graph along with a few examples.
1.  What is a Critical Point of a Function? 
2.  Finding Critical Points 
3.  Critical Points on a Graph 
4.  Critical Points of Multivariable Functions 
5.  FAQs on Critical Points 
What is a Critical Point of a Function?
A critical point of a function y = f(x) is a point (c, f(c)) on the graph of f(x) at which either the derivative is 0 (or) the derivative is not defined. But how is a critical point related to the derivative? We know that the slope of a tangent line of y = f(x) at a point is nothing but the derivative f'(x) at that point. We already have seen that a function has either a horizontal or a vertical tangent at the critical point.
 Horizontal tangent at (c, f(c)) ⇒ Slope = 0 ⇒ f '(c) = 0
 Vertical tangent at (c, f(c)) ⇒ Slope = undefined ⇒ f'(c) is NOT defined
Critical Point of a Function Definition
Based upon the above discussion, a critical point of a function is mathematically defined as follows. A point (c, f(c)) is a critical point of a continuous function y = f(x) if and only if
 c is in the domain of f(x).
 Either f '(c) = 0 or f'(c) is NOT defined.
Critical Values of a Function
The critical values of a function are the values of the function at the critical points. For example, if (c, f(c)) is a critical point of y = f(x) then f(c) is called the critical value of the function corresponding to the critical point (c, f(c)).
Finding Critical Points
Here are the steps to find the critical point(s) of a function based upon the definition. To find the critical point(s) of a function y = f(x):
 Step  1: Find the derivative f '(x).
 Step  2: Set f '(x) = 0 and solve it to find all the values of x (if any) satisfying it.
 Step  3: Find all the values of x (if any) where f '(x) is NOT defined.
 Step  4: All the values of x (only which are in the domain of f(x)) from Step  2 and Step  3 are the xcoordinates of the critical points. To find the corresponding ycoordinates, just substitute each of them into the function y = f(x). Writing all such pairs (x, y) would give us all critical points.
Example to Find Critical Points
Let us find the critical points of the function f(x) = x^{1/3}  x. For this, we first have to find the derivative.
Step  1: f '(x) = (1/3) x^{2/3}  1 = 1 / (3x^{2/3)})  1
Step  2: f'(x) = 0
1 / (3x^{2/3)})  1 = 0
1 / (3x^{2/3)}) = 1
1 = 3x^{2/3}
1/3 = x^{2/3}
Cubing on both sides,
1/27 = x^{2}
Taking square root on both sides,
± 1/(3√3) = x (or) x = ± √3 / 9
So x = √3 / 9 and x =  √3 / 9
Step  3: f'(x) is NOT defined at x = 0.
Step  4: The domain of f(x) is the set of all real numbers and hence all xvalues from Step  2 and Step  3 are present in the domain of f(x) and hence all these are the xcoordinates of the critical points. Let us find their corresponding ycoordinates:
 When x = √3 / 9, y = (√3 / 9)^{1/3}  (√3 / 9) = 2√3 / 9
 When x = √3 / 9, y = (√3 / 9)^{1/3}  (√3 / 9) = 2√3 / 9
 When x = 0, y = 0^{1/3}  0 = 0
Therefore, the critical points of f(x) are (√3 / 9, 2√3 / 9), (√3 / 9, 2√3 / 9) and (0, 0). In this example, the ycoordinates of critical points which are 2√3 / 9, 2√3 / 9, and 0 are the critical values of the function.
Critical Points on a Graph
We have already seen how to find the critical points when a function is given. Now, we will see how to find the critical points from the graph of a function. The following points would help us in identifying the critical points from a given graph.
 We know that the points at which the tangents are horizontal are critical points. So at all such critical points, the graph either changes from "increasing to decreasing" or from "decreasing to increasing". It means the curve may have (but not necessarily) a local maximum or a local minimum at critical points. Here is an example.
In the above figure, (0, 0) and (2, 4) are critical points as we have local minimum and local maximum respectively at these points. Note that we can draw horizontal tangents also at these points.  The points on the curve where we can draw a vertical tangent are also critical points.
In the above figure, (0, 0) is a critical point.  The sharp turning points (cusps) are also critical points.
In the above figure, (0, 0) is a critical point.
Critical Points of Multivariable Functions
For finding the critical points of a singlevariable function y = f(x), we have seen that we set its derivative to zero and solve. But to find the critical points of multivariable functions (functions with more than one variable), we will just set every first partial derivative with respect to each variable to zero and solve the resulting simultaneous equations. For example:
 To find the critical points of a twovariable function f(x, y), set ∂f / ∂x = 0 and ∂f / ∂y = 0 and solve the system of equations.
 To find the critical points of a threevariable function f(x, y, z), set ∂f / ∂x = 0, ∂f / ∂y = 0, and ∂f / ∂z = 0 and solve the resultant system of equations.
Example of Finding Critical Points of a TwoVariable Function
Let us find the critical points of f(x, y) = x^{2} + y^{2} + 2x + 2y. For this, we have to find the partial derivatives first and then set each of them to zero.
∂f / ∂x = 2x + 2 and ∂f / ∂y = 2y + 2
If we set them to zero,
 2x + 2 = 0 ⇒ x = 1
 2y + 2 = 0 ⇒ y = 1
So the critical point is (1, 1).
Important Points on Critical Points:
 The points at which horizontal tangent can be drawn are critical points.
 The points at which vertical tangent can be drawn are critical points.
 All sharp turning points are critical points.
 Local minimum and local maximum points are critical points but a function doesn't need to have a local minimum or local maximum at a critical point. For example, f(x) = 3x^{4}  4x^{3} has critical point at (0, 0) but it is neither a minimum nor a maximum.
 The critical point of a linear function does not exist.
 The critical point of a quadratic function is always its vertex.
Related Topics:
Critical Point Examples

Example 1: Find the critical points of the function f(x) = x^{2/3}.
Solution:
The given function is f(x) = x^{2/3}.
Its derivative is,
f '(x) = (2/3) x^{1/3 }= 2 / (3x^{1/3})
 Setting f'(x) = 0, we get 2 / (3x^{1/3}) = 0 ⇒ 2 = 0, which can never happen. So there are no x values that satisfy f '(x) = 0.
 Now, check where f '(x) is not defined. We can see that 2 / (3x^{1/3}) is NOT defined at x = 0.
So only critical point is at x = 0. Its critical value is, f(0) = 0^{2/3} = 0.
Answer: Critical point = (0, f(0)) = (0, 0).

Example 2: Find the xcoordinates of critical points of f(x) = \(\dfrac{x+3}{x^{2}+3 x+2}\).
Solution:
The given function is, f(x) = \(\dfrac{x+3}{x^{2}+3 x+2}\).
It is defined only when x^{2} + 3x + 2 ≠ 0 ⇒ (x + 1) (x + 2) ≠ 0 ⇒ x ≠ 1 and x ≠ 2.
So its domain is the set of all real numbers except 1 and 2.
Its derivative by using the quotient rule is,
\(\begin{aligned}
f^{\prime}(x) &=\dfrac{\left(x^{2}+3 x+2\right)(1)(x+3)(2 x+3)}{\left(x^{2}+3 x+2\right)^{2}} \\[0.2cm]
&=\dfrac{x^{2}+3 x+22 x^{2}3 x6 x9}{\left(x^{2}+3 x+2\right)^{2}} \\[0.2cm]
&=\dfrac{x^{2}6 x7}{\left(x^{2}+3 x+2\right)^{2}}
\end{aligned}\) Setting f'(x) = 0, we get x^{2}  6x + 7 = 0. Solving this using quadratic formula, we get
\(\begin{aligned}
x &=\frac{(6) \pm \sqrt{(6)^{2}4(1)(7)}}{2(1)} \\[0.2cm]
&=\frac{6 \pm \sqrt{8}}{2} \\[0.2cm]
&=\frac{6 \pm 2 \sqrt{2}}{2} \\[0.2cm]
&=3 \pm \sqrt{2}
\end{aligned}\) We know that f'(x) is NOT defined when its denominator x^{2} + 3x + 2 = 0. By solving this, we get (x + 1) (x + 2) = 0 ⇒ x = 1 and x = 2. But these are NOT critical points as they do NOT lie in the domain of f(x).
Answer: The critical points are at x = 3 + √2 and 3  √2.

Example 3: Find the critical points of the function f(x) = x ln x.
Solution:
The given function is, f(x) = x ln x.
Its domain is (0, ∞).
Its derivative is,
f'(x) = x(1/x) + ln x (1) = 1 + ln x.
 Set f '(x) = 0 ⇒ 1 + ln x = 0 ⇒ ln x = 1 ⇒ x = e^{1} = 1/e.
 There is no x value in the domain of f(x) where f '(x) is NOT defined.
Therefore, there is only one critical value which is at x = 1/e. Its critical value is f(1/e) = (1/e) ln (1/e) = (1/e) (1) = 1/e.
Answer: The critical point of the given function is (1/e, 1/e).
FAQs on Critical Points
What is a Critical Point in Calculus?
A critical point of a function y = f(x) is a point at which the graph of the function is either has a vertical tangent or horizontal tangent. To find critical points we see:
 The points at which f'(x) = 0.
 The points at which f'(x) is NOT defined.
How to Find Critical Points of a Function?
To find the critical points of a function y = f(x), just find xvalues where the derivative f'(x) = 0 and also the xvalues where f'(x) is not defined. These would give the xvalues of the critical points and by substituting each of them in y = f(x) will give the yvalues of the critical points.
How to Find Critical Points On a Graph?
To find the critical points on a graph:
 Check for minimum and maximum points.
 Check the points where drawing a horizontal or vertical tangent is possible.
 Check for sharp turning points (cusps).
How to Find Critical Points of Multivariable Functions?
To find the critical points of a multivariable function, say f(x, y), we just set the partial derivatives with respect to each variable to 0 and solve the equations. i.e., we solve f\(_x\) =0 and f\(_y\) = 0 and solve them.
Is a Critical Point Always a Local Minimum or a Local Maximum?
No, a critical point doesn't need to be a local minimum or local maximum always. For example, the critical point of f(x) = x^{3} is (0, 0) but f(x) neither has a minimum nor a maximum at (0, 0).
What is the Use of Critical Point?
The critical point is used to:
 Find maxima and minima.
 Finding the increasing and decreasing intervals.
 Used in optimization problems.
What are Types of Critical Points?
There can be three types of critical points:
 Critical points where the function has maxima/minima.
 Critical points where there can be a vertical tangent.
 Critical points at which the graph takes a sharp turn.
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