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 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:

    \[{f\left( x \right) = 0,}\;\; \Rightarrow {\left| {{x^2} – 4x + 3} \right| = 0.}\]

    \[{D = {\left( { – 4} \right)^2} – 4 \cdot 3 = 4,}\;\; \Rightarrow {{x_{1,2}} = \frac{{4 \pm \sqrt 4 }}{2} = 1,3.}\]

    critical points of the function f(x)=|x^2-4x+3|
    Figure 8.

    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.

    Critical points of the function f(x)=|x^3-12x|
    Figure 9.

    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).