# Critical Points

• ### 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.

### 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.
3. A critical point $$x = c$$ is a local maximum if the function changes from increasing to decreasing at that point.

4. 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. $f\left( x \right) = 2x – {x^2}.$ ${f^\prime\left( x \right) = \left( {2x – {x^2}} \right)^\prime }={ 2 – 2x.}$ ${f^\prime\left( c \right) = 0,\;\;} \Rightarrow {2 – 2c = 0,\;\;} \Rightarrow {c = 1.}$
5. 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.
6. A critical point $$x = c$$ is an inflection point if the function changes concavity at that point.

7. 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 $$c = 0.$$ $f\left( x \right) = {x^3}.$ $f^\prime\left( x \right) = \left( {{x^3}} \right)^\prime = 3{x^2}.$ ${f^\prime\left( c \right) = 0,\;\;} \Rightarrow {3{c^2} = 0,\;\;} \Rightarrow {c = 0.}$
8. 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 $$f^\prime\left( c \right) = 0.$$ $f\left( x \right) = \left\{ {\begin{array}{*{20}{l}} {2 – x,\;x \le 0}\\ {2,\;x \gt 0} \end{array}} \right..$ $f^\prime\left( x \right) = \left\{ {\begin{array}{*{20}{l}} { – 1,\;x \le 0}\\ {0,\;x \gt 0} \end{array}} \right..$ ${f^\prime\left( c \right) = 0,\;\;} \Rightarrow {c \gt 0.}$

• ## Solved Problems

Click 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.

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

### 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).