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.


1. 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. \[f\left( x \right) = x + {e^{ – x}}.\] \[{f^\prime\left( x \right) = \left( {x + {e^{ – x}}} \right)^\prime }={ 1 – {e^{ – x}}.}\] \[{f^\prime\left( c \right) = 0,\;\;} \Rightarrow {1 – {e^{ – c}} = 0,\;\;} \Rightarrow {{e^{ – c}} = 1,\;\;} \Rightarrow {{e^{ – c}} = {e^0},\;\;} \Rightarrow {c = 0.}\]

   

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.

   

A critical point \(x = c\) is a local maximum if the function changes from increasing to decreasing at that point.


3. 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. \[f\left( x \right) = 2x – {x^2}.\] \[{f^\prime\left( x \right) = \left( {2x – {x^2}} \right)^\prime }={ 2 – 2x.}\] \[{f^\prime\left( c \right) = 0,\;\;} \Rightarrow {2 – 2c = 0,\;\;} \Rightarrow {c = 1.}\]

   

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.

   

A critical point \(x = c\) is an inflection point if the function changes concavity at that point.


5. 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 \(c = 0.\) \[f\left( x \right) = {x^3}.\] \[f^\prime\left( x \right) = \left( {{x^3}} \right)^\prime = 3{x^2}.\] \[{f^\prime\left( c \right) = 0,\;\;} \Rightarrow {3{c^2} = 0,\;\;} \Rightarrow {c = 0.}\]

   

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 \(f^\prime\left( c \right) = 0.\) \[f\left( x \right) = \left\{ {\begin{array}{*{20}{l}} {2 – x,\;x \le 0}\\ {2,\;x \gt 0} \end{array}} \right..\] \[f^\prime\left( x \right) = \left\{ {\begin{array}{*{20}{l}} { – 1,\;x \le 0}\\ {0,\;x \gt 0} \end{array}} \right..\] \[{f^\prime\left( c \right) = 0,\;\;} \Rightarrow {c \gt 0.}\]

   

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Solved Problems

Click or tap a problem to see the solution.

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.

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 \({c_1} = 0,{c_2} = 6.\)

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:

\[{c_1} = 0,\,{c_2} = 1,\,{c_3} = 3.\]

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.

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:

\[{{c_1} = – \frac{{\sqrt 2 }}{2},}\;{{c_2} = \frac{{\sqrt 2 }}{2}.}\]

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 \(c = 1.\)

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:

\[{{c_1} = – \sqrt 5,}\;{{c_2} = 0,}\;{{c_3} = \sqrt 5 .}\]

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 \(c = e.\)

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?)

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:

\[{{c_1} = \pi ,}\;{{c_2} = \frac{{2\pi }}{3}.}\]

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).