Calculus

Applications of the Derivative

Applications of Derivative Logo

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.

Classification of Critical Points

Types of critical points
Figure 1.

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.
    \[f\left( x \right) = x + {e^{ - x}}.\]
    \[f^\prime\left( x \right) = \left( {x + {e^{ - x}}} \right)^\prime = 1 - {e^{ - x}}.\]
    \[f^\prime\left( c \right) = 0,\;\; \Rightarrow 1 - {e^{ - c}} = 0,\;\; \Rightarrow {e^{ - c}} = 1,\;\; \Rightarrow {e^{ - c}} = {e^0},\;\; \Rightarrow c = 0.\]
    Local minimum U-point
    Figure 2.
  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.
    Local minimum V-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.
    \[f\left( x \right) = 2x - {x^2}.\]
    \[f^\prime\left( x \right) = \left( {2x - {x^2}} \right)^\prime = 2 - 2x.\]
    \[f^\prime\left( c \right) = 0,\;\; \Rightarrow 2 - 2c = 0,\;\; \Rightarrow c = 1.\]
    Local maximum U-point
    Figure 4.
  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.
    Local maximum V-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 \(c = 0.\)
    \[f\left( x \right) = {x^3}.\]
    \[f^\prime\left( x \right) = \left( {{x^3}} \right)^\prime = 3{x^2}.\]
    \[f^\prime\left( c \right) = 0,\;\; \Rightarrow 3{c^2} = 0,\;\; \Rightarrow c = 0.\]
    Inflection point is a critical point
    Figure 6.
  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 \(f^\prime\left( c \right) = 0.\)
    \[f\left( x \right) = \left\{ {\begin{array}{*{20}{l}} {2 - x,\;x \le 0}\\ {2,\;x \gt 0} \end{array}} \right..\]
    \[f^\prime\left( x \right) = \left\{ {\begin{array}{*{20}{l}} { - 1,\;x \le 0}\\ {0,\;x \gt 0} \end{array}} \right..\]
    \[f^\prime\left( c \right) = 0,\;\; \Rightarrow c \gt 0.\]
    Critical points of a constant function
    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 \[f\left( x \right) = \frac{{{x^2} - 4x + 3}}{{x - 2}}.\]

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.

Answer:

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

Example 6.

Indicate all critical points of the function \[f\left( x \right) = \frac{{{e^x}}}{x}.\]

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.

Page 1 Page 2