# Calculus

## Applications of the Derivative # 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. Figure 3.
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. Figure 5.
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 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 $f\left( x \right) = \frac{{{x^2} - 4x + 3}}{{x - 2}}.$

### 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 $f\left( x \right) = \frac{{{e^x}}}{x}.$

### 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:

$f^\prime\left( c \right) = 0,\;\; \Rightarrow c\left( {2\ln c + 1} \right) = 0.$

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 $${c_2} = \frac{1}{{\sqrt e }}$$ 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 $f\left( x \right) = \frac{{{x^2} - 4x + 3}}{{x - 2}}.$

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 $f\left( x \right) = \frac{{{e^x}}}{x}.$

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 $$c = 1.$$

See more problems on Page 2.