Definition
Let \(f\left(x\right)\) be a function and let \(c\) be a point in the domain of the function. The point \(c\) is called a critical point of \(f\) if either \(f’\left( c \right) = 0\) or \(f’\left( c \right)\) does not exist.
Classification of Critical Points
Examples of Critical Points
-
A critical point \(x = c\) is a local minimum if the function changes from decreasing to increasing at that point.
- The function \(f\left( x \right) = x + {e^{ – x}}\) has a critical point (local minimum) at \(c = 0.\) The derivative is zero at this point.
\[f\left( x \right) = x + {e^{ – x}}.\]
\[{f^\prime\left( x \right) = \left( {x + {e^{ – x}}} \right)^\prime }={ 1 – {e^{ – x}}.}\]
\[{f^\prime\left( c \right) = 0,\;\;} \Rightarrow {1 – {e^{ – c}} = 0,\;\;} \Rightarrow {{e^{ – c}} = 1,\;\;} \Rightarrow {{e^{ – c}} = {e^0},\;\;} \Rightarrow {c = 0.}\]
Figure 2. - The function \(f\left( x \right) = \left| {x – 3} \right|\) has a critical point (local minimum) at \(c = 3.\) The derivative does not exist at this point.
Figure 3.
A critical point \(x = c\) is a local maximum if the function changes from increasing to decreasing at that point.
- The function \(f\left( x \right) = 2x – {x^2}\) has a critical point (local maximum) at \(c = 1.\) The derivative is zero at this point.
\[f\left( x \right) = 2x – {x^2}.\]
\[{f^\prime\left( x \right) = \left( {2x – {x^2}} \right)^\prime }={ 2 – 2x.}\]
\[{f^\prime\left( c \right) = 0,\;\;} \Rightarrow {2 – 2c = 0,\;\;} \Rightarrow {c = 1.}\]
Figure 4. - The function \(f\left( x \right) = 1 – \left| {x + 2} \right|\) has a critical point (local maximum) at \(c = -2.\) The derivative does not exist at this point.
Figure 5.
A critical point \(x = c\) is an inflection point if the function changes concavity at that point.
- The function \(f\left( x \right) = {x^3}\) has a critical point (inflection point) at \(c = 0.\) The first and second derivatives are zero at \(c = 0.\)
\[f\left( x \right) = {x^3}.\]
\[f^\prime\left( x \right) = \left( {{x^3}} \right)^\prime = 3{x^2}.\]
\[{f^\prime\left( c \right) = 0,\;\;} \Rightarrow {3{c^2} = 0,\;\;} \Rightarrow {c = 0.}\]
Figure 6. - Trivial case: Each point of a constant function is critical. For example, any point \(c \gt 0\) of the function \(f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}
{2 – x,\;x \le 0}\\
{2,\;x \gt 0}
\end{array}} \right.\) is a critical point since \(f^\prime\left( c \right) = 0.\)
\[f\left( x \right) = \left\{ {\begin{array}{*{20}{l}}
{2 – x,\;x \le 0}\\
{2,\;x \gt 0}
\end{array}} \right..\]
\[f^\prime\left( x \right) = \left\{ {\begin{array}{*{20}{l}}
{ – 1,\;x \le 0}\\
{0,\;x \gt 0}
\end{array}} \right..\]
\[{f^\prime\left( c \right) = 0,\;\;} \Rightarrow {c \gt 0.}\]
Figure 7.
Solved Problems
Click or tap a problem to see the solution.
Example 1
Find the critical points of the function \(f\left( x \right) = {x^2}\ln x\).Example 2
Find the critical points of the function \(f\left( x \right) = 8{x^3} – {x^4}.\)Example 3
Find the critical points of the function \(f\left( x \right) = {x^4} – 5{x^4} + 5{x^3} – 1.\)Example 4
Find the critical points of the function \(f\left( x \right) = \large{\frac{{{x^2} – 4x + 3}}{{x – 2}}}\normalsize.\)Example 5
Find all critical points of the function \(f\left( x \right) = x\sqrt {1 – {x^2}}.\)Example 6
Indicate all critical points of the function \(f\left( x \right) = \large{\frac{{{e^x}}}{x}}\normalsize.\)Example 7
Indicate all critical points of the function \(f\left( x \right) = \left| {{x^2} – 5} \right|.\)Example 8
Find all critical points of the function \(f\left( x \right) = \large{\frac{x}{{\ln x}}}\normalsize.\)Example 9
Determine critical points of the function \(f\left( x \right) = \left| {{x^2} – 4x + 3} \right|.\)Example 10
Find the critical points of the function \(f\left( x \right) = {\sin ^2}x – \cos x\) on the interval \(\left( {0,2\pi } \right).\)Example 11
How many critical points does the function \(f\left( x \right) = \left| {{x^3} – 12x} \right|\) have?Example 1.
Find the critical points of the function \(f\left( x \right) = {x^2}\ln x\).Solution.
Take the derivative using the product rule:
\[{f^\prime\left( x \right) = \left( {{x^2}\ln x} \right)^\prime }={ 2x \cdot \ln x + {x^2} \cdot \frac{1}{x} }={ 2x\ln x + x }={ x\left( {2\ln x + 1} \right).}\]
Determine the points where the derivative is zero:
\[{f^\prime\left( c \right) = 0,\;\;} \Rightarrow \cssId{element14}{c\left( {2\ln c + 1} \right) = 0.}\]
The first root \({c_1} = 0\) is not a critical point because the function is defined only for \(x \gt 0.\)
Consider the second root:
\[{2\ln c + 1 = 0,\;\;} \Rightarrow {\ln c = – \frac{1}{2},\;\;} \Rightarrow {{c_2} = {e^{ – \frac{1}{2}}} = \frac{1}{{\sqrt e }}.}\]
Hence \({c_2} = \large{\frac{1}{{\sqrt e }}}\normalsize\) is a critical point of the given function.
Example 2.
Find the critical points of the function \(f\left( x \right) = 8{x^3} – {x^4}.\)Solution.
The function is defined and differentiable over the entire set of real numbers. Therefore, we just differentiate it to determine where the derivative is zero.
\[{f^\prime\left( x \right) = \left( {8{x^3} – {x^4}} \right)^\prime }={ 24{x^2} – 4{x^3}.}\]
\[{f^\prime\left( c \right) = 0,}\;\; \Rightarrow {24{c^2} – 4{c^3} = 0,}\;\; \Rightarrow {4{c^2}\left( {6 – c} \right) = 0,}\;\; \Rightarrow {{c_1} = 0,{c_2} = 6.}\]
Hence, the function has 2 critical points \({c_1} = 0,{c_2} = 6.\)
Example 3.
Find the critical points of the function \(f\left( x \right) = {x^4} – 5{x^4} + 5{x^3} – 1.\)Solution.
The function is defined and differentiable for all \(x\). Calculate the derivative:
\[{f^\prime\left( x \right) = \left( {{x^4} – 5{x^4} + 5{x^3} – 1} \right)^\prime }={ 5{x^4} – 20{x^3} + 15{x^2}.}\]
By equating the derivative to zero, we get the critical points:
\[{f^\prime\left( c \right) = 0,}\;\; \Rightarrow {5{c^4} – 20{c^3} + 15{c^2} = 0,}\;\; \Rightarrow {5{c^2}\left( {{c^2} – 4c + 3} \right) = 0.}\]
We can factor the quadratic expression:
\[{c^2} – 4c + 3 = \left( {c – 1} \right)\left( {c – 3} \right),\]
so the latter equation is written as
\[5{c^2}\left( {c – 1} \right)\left( {c – 3} \right) = 0.\]
Thus, the function has the following critical points:
\[{c_1} = 0,\,{c_2} = 1,\,{c_3} = 3.\]
Example 4.
Find the critical points of the function \(f\left( x \right) = \large{\frac{{{x^2} – 4x + 3}}{{x – 2}}}\normalsize.\)Solution.
Take the derivative by the quotient rule:
\[{f^\prime\left( x \right) = \left( {\frac{{{x^2} – 4x + 3}}{{x – 2}}} \right)^\prime }={ \frac{{\left( {2x – 4} \right)\left( {x – 2} \right) – \left( {{x^2} – 4x + 3} \right) \cdot 1}}{{{{\left( {x – 2} \right)}^2}}} = \frac{{{x^2} – 4x + 5}}{{{{\left( {x – 2} \right)}^2}}}.}\]
Solve the equation \(f^\prime\left( c \right) = 0:\)
\[{\frac{{{c^2} – 4c + 5}}{{{{\left( {c – 2} \right)}^2}}} = 0,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {{c^2} – 4c + 5 = 0}\\ {c – 2 \ne 0} \end{array}} \right..}\]
The quadratic equation has no roots as the discriminant \(D = 16 – 20 = – 4 \lt 0.\)
Note that \(x = 2\) is a not a critical point as the function is not defined at this point.
Thus, the given function has no critical points.
Example 5.
Find all critical points of the function \(f\left( x \right) = x\sqrt {1 – {x^2}}.\)Solution.
First we determine the domain of the function:
\[{1 – {x^2} \ge 0,}\;\; \Rightarrow {{x^2} \le 1,}\;\; \Rightarrow {- 1 \le x \le 1.}\]
Find the derivative by the product rule:
\[{f^\prime\left( x \right) = \left( {x\sqrt {1 – {x^2}} } \right)^\prime }={ x^\prime\sqrt {1 – {x^2}} + x\left( {\sqrt {1 – {x^2}} } \right)^\prime }={ \sqrt {1 – {x^2}} + x \cdot \frac{{\left( { – 2x} \right)}}{{2\sqrt {1 – {x^2}} }} }={ \frac{{1 – {x^2} – {x^2}}}{{\sqrt {1 – {x^2}} }} }={ \frac{{1 – 2{x^2}}}{{\sqrt {1 – {x^2}} }}.}\]
Determine the points at which the derivative is zero:
\[{f^\prime\left( c \right) = 0,}\;\; \Rightarrow {\frac{{1 – 2{c^2}}}{{\sqrt {1 – {c^2}} }} = 0,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {1 – 2{c^2} = 0}\\ {\sqrt {1 – {c^2}} \ne 0} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {{c^2} = \frac{1}{2}}\\ {{c^2} \ne 1} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {{c_{1,2}} = \pm \frac{{\sqrt 2 }}{2}}\\ {c \ne \pm 1} \end{array}} \right.,}\;\; \Rightarrow {{c_{1,2}} = \pm \frac{{\sqrt 2 }}{2}.}\]
So we have two points in the domain of the function where the derivative is zero. Hence, these points are critical, by definition.
Answer:
\[{{c_1} = – \frac{{\sqrt 2 }}{2},}\;{{c_2} = \frac{{\sqrt 2 }}{2}.}\]
Example 6.
Indicate all critical points of the function \(f\left( x \right) = \large{\frac{{{e^x}}}{x}}\normalsize.\)Solution.
The function is defined over all \(x\) except \(x = 0\) where it has a discontinuity. Differentiate the function using the quotient rule:
\[{f^\prime\left( x \right) = \left( {\frac{{{e^x}}}{x}} \right)^\prime }={ \frac{{\left( {{e^x}} \right)^\prime \cdot x – {e^x} \cdot x^\prime}}{{{x^2}}} }={ \frac{{{e^x} \cdot x – {e^x} \cdot 1}}{{{x^2}}} }={ \frac{{\left( {x – 1} \right){e^x}}}{{{x^2}}}.}\]
Solve the equation \(f^\prime\left( c \right) = 0:\)
\[{f^\prime\left( c \right) = 0,}\;\;\Rightarrow {\frac{{\left( {c – 1} \right){e^c}}}{{{c^2}}} = 0,}\;\; \Rightarrow {c = 1.}\]
Note that \(c =0\) is not a critical point since the function itself is not defined here.
Therefore, the function has one critical point \(c = 1.\)
Example 7.
Indicate all critical points of the function \(f\left( x \right) = \left| {{x^2} – 5} \right|.\)Solution.
Find the roots of the function:
\[{f\left( x \right) = 0,}\;\; \Rightarrow {\left| {{x^2} – 5} \right| = 0,}\;\; \Rightarrow {{x_{1,2}} = \pm \sqrt 5 .}\]
The derivative does not exist at the corner points \(x = – \sqrt 5 \) and \(x = \sqrt 5 ,\) i.e. these points are critical.
In the interval \(\left[ { – \sqrt 5 ,\sqrt 5 } \right],\) the function is written as
\[{f\left( x \right) = – \left( {{x^2} – 5} \right) }={ – {x^2} + 5.}\]
Solving the equation \(f^\prime\left( c \right) = 0\) on this interval, we get one more critical point:
\[{f^\prime\left( c \right) = 0,}\;\; \Rightarrow {- 2c = 0,}\;\; \Rightarrow {c = 0.}\]
Hence, the function has three critical points:
\[{{c_1} = – \sqrt 5,}\;{{c_2} = 0,}\;{{c_3} = \sqrt 5 .}\]
Example 8.
Find all critical points of the function \(f\left( x \right) = \large{\frac{x}{{\ln x}}}\normalsize.\)Solution.
The domain of \(f\left( x \right)\) is determined by the conditions:
\[\left\{ \begin{array}{l} x \gt 0\\ \ln x \ne 0 \end{array} \right., \Rightarrow \left\{ \begin{array}{l} x \gt 0\\ x \ne 1 \end{array} \right..\]
Take the derivative using the quotient rule:
\[{f^\prime\left( x \right) = \left( {\frac{x}{{\ln x}}} \right)^\prime }={ \frac{{x^\prime \cdot \ln x – x \cdot \left( {\ln x} \right)^\prime}}{{{{\ln }^2}x}} }={ \frac{{1 \cdot \ln x – x \cdot \frac{1}{x}}}{{{{\ln }^2}x}} }={ \frac{{\ln x – 1}}{{{{\ln }^2}x}}.}\]
Equating the derivative to zero, we find the critical points \(c:\)
\[{f^\prime\left( c \right) = 0,}\;\; \Rightarrow {\frac{{\ln c – 1}}{{{{\ln }^2}c}} = 0,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {\ln c = 1}\\ {{{\ln }^2}c \ne 0} \end{array}} \right.,}\;\; \Rightarrow {\left\{ {\begin{array}{*{20}{l}} {c = e}\\ {c \ne 1} \end{array}} \right..}\]
Note that the derivative does not exist at \(c = 1\) (where the denominator of the derivative approaches zero). But the function itself is also undefined at this point. Therefore, \(c = 1\) is not a critical point.
Hence, the function has one critical point \(c = e.\)
Example 9.
Determine critical points of the function \(f\left( x \right) = \left| {{x^2} – 4x + 3} \right|.\)Solution.
First, we find the roots of the function and sketch its graph:
\[{f\left( x \right) = 0,}\;\; \Rightarrow {\left| {{x^2} – 4x + 3} \right| = 0.}\]
\[{D = {\left( { – 4} \right)^2} – 4 \cdot 3 = 4,}\;\; \Rightarrow {{x_{1,2}} = \frac{{4 \pm \sqrt 4 }}{2} = 1,3.}\]
We see that the function has two corner points (or V-points): \(c = 1\) and \(c = 3,\) where the derivative does not exist. Therefore, \(c = 1\) and \(c = 3\) are critical points of the function.
Besides that, the function has one more critical point at which the derivative is zero. Indeed, in the interval \(1 \le x \le 3,\) the function is written as
\[{f\left( x \right) = – \left( {{x^2} – 4x + 3} \right) }={ – {x^2} + 4x – 3.}\]
Differentiating and equating to zero, we get
\[{f^\prime\left( x \right) = \left( { – {x^2} + 4x – 3} \right)^\prime }={ – 2x + 4.}\]
\[{f^\prime\left( c \right) = 0,}\;\; \Rightarrow {- 2c + 4 = 0,}\;\; \Rightarrow {c = 2.}\]
Thus, the function has three critical points:
\[{{c_1} = 1,}\;{{c_2} = 2,}\;{{c_3} = 3.}\]
(a pretty nice answer, isn’t it?)
Example 10.
Find the critical points of the function \(f\left( x \right) = {\sin ^2}x – \cos x\) on the interval \(\left( {0,2\pi } \right).\)Solution.
Determine the derivative of \(f\left( x \right)\) using the chain rule and trig derivatives:
\[{f^\prime\left( x \right) = \left( {{{\sin }^2}x – \cos x} \right)^\prime }={ 2\sin x\cos x – \left( { – \sin x} \right) }={ 2\sin x\cos x + \sin x }={ \sin x\left( {2\cos x + 1} \right).}\]
Solving the equation \(f^\prime\left( c \right) = 0,\) we obtain two solutions:
\[{f^\prime\left( c \right) = 0,}\;\; \Rightarrow {\sin c\left( {2\cos c + 1} \right) = 0.}\]
\[{1.\;\sin c = 0,}\;\; \Rightarrow {c = \pi n,\;n \in Z.}\]
The equation \(\sin c = 0\) has one root \(c = \pi\) in the open interval \(\left( {0,2\pi } \right).\)
\[{2.\;2\cos c + 1 = 0,}\;\; \Rightarrow {2\cos x = – 1,}\;\; \Rightarrow {\cos c = – \frac{1}{2},}\;\; \Rightarrow {c = \pm \arccos \left( { – \frac{1}{2}} \right) + 2\pi n,}\;\; \Rightarrow {c = \pm \frac{{2\pi }}{3} + 2\pi n,\,n \in Z.}\]
Only one solution \(c = \large{\frac{{2\pi }}{3}}\normalsize\) belongs to the open interval \(\left( {0,2\pi } \right).\)
So, the function has two critical points:
\[{{c_1} = \pi ,}\;{{c_2} = \frac{{2\pi }}{3}.}\]
Example 11.
How many critical points does the function \(f\left( x \right) = \left| {{x^3} – 12x} \right|\) have?Solution.
Determine the roots of the function:
\[{f\left( x \right) = 0,}\;\; \Rightarrow {\left| {{x^3} – 12x} \right| = 0,}\;\; \Rightarrow {x\left( {{x^2} – 12} \right) = 0,}\;\; \Rightarrow {{x_1} = 0,\,{x_{2,3}} = \pm 2\sqrt 3 .}\]
We see that the function has 3 corner points (or V-points) at \(x = – 2\sqrt 3 ,\) \(x = 0\) and \(x = 2\sqrt 3 .\) Since the derivative does not exist at these points, we have 3 critical points here.
Consider other critical points which can occur at local extrema.
In the interval \(\left[ { – 2\sqrt 3 ,0} \right],\) the function has the form
\[f\left( x \right) = {x^3} – 12x.\]
Its derivative is given by
\[{f^\prime\left( x \right) = \left( {{x^3} – 12x} \right)^\prime }={ 3{x^2} – 12.}\]
By equating the derivative to zero, we get
\[{f^\prime\left( x \right) = 0,}\;\; \Rightarrow {3{x^2} – 12 = 0,}\;\; \Rightarrow {x = \pm 2.}\]
Only one point \(x = -2\) lies in the interval under consideration. So \(x = -2\) is also a critical point.
Similarly, we find that the function has one more critical point \(x = 2\) in the interval \(\left[ {0,2\sqrt 3 } \right]\).
Hence, the function has \(5\) critical points (\(3\) V-points and \(2\) local extrema points).