Calculus

Applications of the Derivative

Applications of Derivative Logo

Inflection Points

  • Definition of an Inflection Point

    Consider a function \(y = f\left( x \right),\) which is continuous at a point \({x_0}.\) The function \(f\left( x \right)\) can have a finite or infinite derivative \(f’\left( {{x_0}} \right)\) at this point. If, when passing through \({x_0}\), the function changes the direction of convexity, i.e. there exists a number \(\delta \gt 0\) such that the function is convex upward on one of the intervals \(\left( {{x_0} – \delta ,{x_0}} \right)\) or \(\left( {{x_0},{x_0} + \delta } \right)\), and is convex downward on the other, then \({x_0}\) is called a point of inflection of the function \(y = f\left( x \right).\)

    The geometric meaning of an inflection point is that the graph of the function \(f\left( x \right)\) passes from one side of the tangent line to the other at this point, i.e. the curve and the tangent line intersect (see Figure \(1\)).

    The geometric meaning of an inflection point
    Figure 1.

    Another interesting feature of an inflection point is that the graph of the function \(f\left( x \right)\) in the vicinity of the inflection point \({x_0}\) is located within a pair of the vertical angles formed by the tangent and normal (Figure \(2\)).

    The graph of a function at inflection point and position of the tangent and normal
    Figure 2.

    Necessary Condition for an Inflection Point (Second Derivative Test)

    If \({x_0}\) is a point of inflection of the function \(f\left( x \right)\), and this function has a second derivative in some neighborhood of \({x_0},\) which is continuous at the point \({x_0}\) itself, then

    \[f^{\prime\prime}\left( {{x_0}} \right) = 0.\]

    Proof.

    Suppose that the second derivative at the inflection point \({x_0}\) is not zero: \(f^{\prime\prime}\left( {{x_0}} \right) \ne 0.\) Since it is continuous at \({x_0},\) then there exists a \(\delta\)-neighborhood of the point \({x_0}\) where the second derivative preserves its sign, that is

    \[{f^{\prime\prime}\left( {{x_0}} \right) \lt 0\;\;\text{or}\;\;\;}\kern-0.3pt{f^{\prime\prime}\left( {{x_0}} \right) \lt 0\;\forall \;x \in \left( {{x_0} – \delta ,{x_0} + \delta } \right).}\]

    In this case, the function is either strictly convex upward (when \(f^{\prime\prime}\left( x \right) \lt 0\)) or strictly convex downward (when \(f^{\prime\prime}\left( x \right) \gt 0\)). But then the point \({x_0}\) is not an inflection point. Hence, the assumption is wrong and the second derivative of the inflection point must be equal to zero.

    First Sufficient Condition for an Inflection Point (Second Derivative Test)

    If the function \(f\left( x \right)\) is continuous and differentiable at a point \({x_0},\) has a second derivative \(f^{\prime\prime}\left( {{x_0}} \right)\) in some deleted \(\delta\)-neighborhood of the point \({x_0}\) and if the second derivative changes sign when passing through the point \({x_0},\) then \({x_0}\) is a point of inflection of the function \(f\left( x \right).\)

    Proof.

    Suppose, for example, that the second derivative \(f^{\prime\prime}\left( x \right)\) changes sign from plus to minus when passing through the point \({x_0}.\) Hence, in the left \(\delta\)-neighborhood \(\left( {{x_0} – \delta ,{x_0}} \right),\) the inequality \(f^{\prime\prime}\left( x \right) \gt 0,\) holds, and in the right \(\delta\)-neighborhood \(\left( {{x_0},{x_0} + \delta } \right),\) the inequality \(f^{\prime\prime}\left( x \right) \lt 0\) is valid.

    In this case, according to the sufficient conditions for convexity, the function \(f\left( x \right)\) is convex downward in the left \(\delta\)-neighborhood of the point \({x_0}\) and is convex upward in the right \(\delta\)-neighborhood.

    Consequently, the function changes the direction of convexity at the point \({x_0},\) that is by definition, \({x_0}\) is a point of inflection.

    Second Sufficient Condition for an Inflection Point (Third Derivative Test)

    Let \(f^{\prime\prime}\left( {{x_0}} \right) = 0,\) \(f^{\prime\prime\prime}\left( {{x_0}} \right) \ne 0.\) Then \({x_0}\) is a point of inflection of the function \(f\left( x \right).\)

    Proof.

    As \(f^{\prime\prime\prime}\left( {{x_0}} \right) \ne 0,\) the second derivative is either strictly increasing at \({x_0}\) (if \(f^{\prime\prime\prime}\left( {{x_0}} \right) \gt 0\)) or strictly decreasing at this point (if \(f^{\prime\prime\prime}\left( {{x_0}} \right) \lt 0\)). Because \(f^{\prime\prime}\left( {{x_0}} \right) = 0,\) then the second derivative for some \(\delta \gt 0\) has different signs in the left and right \(\delta\)-neighborhood of \({x_0}.\) Hence, on the basis of the previous theorem, it follows that \({x_0}\) is a point of inflection of the function \(f\left( x \right).\)


  • Solved Problems

    Example 1

    Determine whether the point \(x = 0\) is an inflection point of the function \(f\left( x \right) = \sin x.\)

    Example 2

    Find the points of inflection of the function \(f\left( x \right) = {x^3} – 3{x^2} – 1.\)

    Example 3

    Find the points of inflection of the function \(f\left( x \right) = 2{x^3} + 6{x^2} – 5x + 1.\)

    Example 4

    Find the inflection points of the function \(f\left( x \right) = {x^4} – 6{x^2}.\)

    Example 5

    Find the points of inflection of the function \(f\left( x \right) = 2{x^4} – 3x.\)

    Example 6

    Find the points of inflection of the function \[{f\left( x \right) \text{ = }}\kern0pt{ {x^4} – 12{x^3} + 48{x^2} + 12x + 1.}\]

    Example 7

    Find the inflection points of the function \(f\left( x \right) = 3{x^5} + 5{x^4} – 20{x^3}.\)

    Example 8

    Find the points of inflection of the function \(f\left( x \right) = {x^2}\ln x.\)

    Example 9

    Find the points of inflection of the function \(f\left( x \right) = x{e^{ – 2x}}.\)

    Example 10

    Find the points of inflection of the function \(f\left( x \right) = {e^{ – {x^2}}}.\)

    Example 11

    Find the inflection points of the function \(f\left( x \right) = x + {x^{\large{\frac{5}{3}}\normalsize}}.\)

    Example 12

    Find the points of inflection of the function \(f\left( x \right) = {e^{\sin x}}.\)

    Example 13

    For what values of \(a\) and \(b\) the point \(\left( { – 1,2} \right)\) is an inflection point of the graph of the function \(y\left( x \right) = a{x^3} + b{x^2}?\)

    Example 14

    Find the points of inflection of the function \(f\left( x \right) = \large{\frac{{{x^3}}}{{1 – {x^2}}}}\normalsize.\)

    Example 15

    Find the inflection points of the function \(f\left( x \right) = {x^2} – \large{\frac{1}{{{x^2}}}}\normalsize.\)

    Example 16

    Find the points of inflection of the function \(f\left( x \right) = \large{\frac{{{x^2}}}{{1 + {x^2}}}}\normalsize.\)

    Example 17

    Find the inflection points of a Gaussian function.

    Example 18

    Find the points of inflection of the function \[f\left( x \right) = \sqrt[\large 3\normalsize]{{{x^2}\left( {x + 1} \right)}}.\]

    Example 19

    Find the inflection points of the curve defined by the parametric equations: \[x = {t^2},\;\;y = t + {t^3}.\]

    Example 20

    Show that the graph of the function \(y = {\large\frac{{x + 1}}{{{x^2} + 1}}\normalsize}\) has three points of inflection lying on one straight line.

    Example 1.

    Determine whether the point \(x = 0\) is an inflection point of the function \(f\left( x \right) = \sin x.\)

    Solution.

    We use the second sufficient condition for the existence of an inflection point. Find the derivatives of the sine function up to the third order:

    \[
    {f’\left( x \right) = {\left( {\sin x} \right)^\prime } = \cos x,\;\;\;}\kern-0.3pt
    {f^{\prime\prime}\left( x \right) = {\left( {\cos x} \right)^\prime } = – \sin x,\;\;\;}\kern-0.3pt
    {f^{\prime\prime\prime}\left( x \right) = {\left( { – \sin x} \right)^\prime } = – \cos x.}
    \]

    At the point \(x = 0,\) the second and third derivatives have the following values:

    \[
    {f^{\prime\prime}\left( 0 \right) = – \sin 0 = 0,\;\;\;}\kern-0.3pt
    {f^{\prime\prime\prime}\left( 0 \right) = – \cos 0 = – 1.}
    \]

    Thus, we have \(f^{\prime\prime}\left( {{x_0}} \right) = 0,\) \(f^{\prime\prime\prime}\left( {{x_0}} \right) \ne 0.\) Hence, by the second sufficient condition, the point \(x = 0\) is a point of inflection.

    Example 2.

    Find the points of inflection of the function \(f\left( x \right) = {x^3} – 3{x^2} – 1.\)

    Solution.

    Sign chart for the second derivative of f(x)=x^3-3x^2-1
    Figure 3.

    Compute the first and second derivatives:

    \[{f^\prime\left( x \right) }={ \left( {{x^3} – 3{x^2} – 1} \right)^\prime }={ 3{x^2} – 6x;}\]

    \[{f^{\prime\prime}\left( x \right) }={ \left( {3{x^2} – 6x} \right)^\prime }={ 6x – 6.}\]

    We see that \(f^{\prime\prime}\left( x \right) = 0\) at \(x = 1.\) The function changes concavity as shown in figure above. Since

    \[{f\left( 1 \right) = {1^3} – 3 \cdot {1^2} – 1 }={ – 3,}\]

    the inflection point is at \(\left( {1, – 3} \right).\)

    Example 3.

    Find the points of inflection of the function \(f\left( x \right) = 2{x^3} + 6{x^2} – 5x + 1.\)

    Solution.

    We differentiate this function twice to get the second derivative:

    \[{f^\prime\left( x \right) }={ \left( {2{x^3} + 6{x^2} – 5x + 1} \right)^\prime }={ 6{x^2} + 12x – 5;}\]

    \[{f^{\prime\prime}\left( x \right) }={ \left( {6{x^2} + 12x – 5} \right)^\prime }={ 12x + 12.}\]

    Clearly that \(f^{\prime\prime}\left( x \right)\) exists for all \(x.\) Determine the points where it is equal to zero:

    \[{f^{\prime\prime}\left( x \right) = 0,}\;\; \Rightarrow {12x + 12 = 0,}\;\; \Rightarrow {x = – 1.}\]

    The function \(f\left( x \right)\) is concave down \(\left( {f^{\prime\prime} \lt 0} \right)\) for \(x \lt -1\) and it is concave up \(\left( {f^{\prime\prime} \gt 0} \right)\) for \(x \gt -1.\) Therefore, \(x = -1\) is an inflection point.

    Calculate the corresponding \(y-\)coordinate:

    \[{f\left( { – 1} \right) }={ 2 \cdot {\left( { – 1} \right)^3} + 6 \cdot {\left( { – 1} \right)^2} }-{ 5 \cdot \left( { – 1} \right) + 1 }={ – 2 + 6 + 5 + 1 }={ 10.}\]

    So, the inflection point is \(\left( {-1, 10} \right).\)

    Example 4.

    Find the inflection points of the function \(f\left( x \right) = {x^4} – 6{x^2}.\)

    Solution.

    Sign chart for the second derivative of f(x)=x^4-6x^2
    Figure 4.

    Compute the first derivative:

    \[{f^\prime\left( x \right) = \left( {{x^4} – 6{x^2}} \right)^\prime }={ 4{x^3} – 12x.}\]

    The second derivative is

    \[{f^{\prime\prime}\left( x \right) = \left( {4{x^3} – 12x} \right)^\prime }={ 12{x^2} – 12 }={ 12\left( {{x^2} – 1} \right).}\]

    Find the roots of the second derivative:

    \[{f^{\prime\prime}\left( x \right) = 0,}\;\; \Rightarrow {12\left( {{x^2} – 1} \right) = 0,}\;\; \Rightarrow {{x_1} = – 1,{x_2} = 1.}\]

    We need to determine where the second derivative changes sign. Draw a sign chart for \(f^{\prime\prime}\left( x \right)\) (see above).

    Clearly, the concavity changes at both points, \(x = -1\) and \(x = 1.\) Hence, these points are points of inflection.

    We can easily calculate their \(y-\)coordinates:

    \[{f\left( { – 1} \right) }={ {\left( { – 1} \right)^4} – 6{\left( { – 1} \right)^2} }={ – 5;}\]

    \[{f\left( 1 \right) }={ {1^4} – 6 \cdot {1^2} }={ – 5.}\]

    So, the inflection points are \(\left( {-1,-5} \right)\) and \(\left( {1,-5} \right).\)

    Example 5.

    Find the points of inflection of the function \(f\left( x \right) = 2{x^4} – 3x.\)

    Solution.

    Sign chart for the second derivative of f(x)=2x^4-3x
    Figure 5.

    We compute the derivatives and draw a sign chart for the second derivative.

    \[{f^\prime\left( x \right) }={ \left( {2{x^4} – 3x} \right)^\prime }={ 8{x^3} – 3;}\]

    \[{f^{\prime\prime}\left( x \right) }={ \left( {8{x^3} – 3} \right)^\prime }={ 24{x^2}.}\]

    We see that the concavity does not change at \(x = 0.\) Consequently, \(x = 0\) is not a point of inflection.

    The second derivative is a continuous function defined over all \(x\). Therefore, we conclude that \(f\left( x \right)\) has no inflection points.

    Example 6.

    Find the points of inflection of the function \[{f\left( x \right) \text{ = }}\kern0pt{ {x^4} – 12{x^3} + 48{x^2} + 12x + 1.}\]

    Solution.

    Find the derivatives:

    \[
    {f’\left( x \right) }
    = {{\left( {{x^4} – 12{x^3} + 48{x^2} + 12x + 1} \right)^\prime } }
    = {4{x^3} – 36{x^2} + 96x + 12 }
    = {4\left( {{x^3} – 9{x^2} + 24x + 3} \right);}
    \]

    \[
    {f^{\prime\prime}\left( x \right) }
    = {{\left( {4\left( {{x^3} – 9{x^2} + 24x + 3} \right)} \right)^\prime } }
    = {4\left( {3{x^2} – 18x + 24} \right) }
    = {12\left( {{x^2} – 6x + 8} \right).}
    \]

    Calculate the roots of the second derivative:

    \[
    {f^{\prime\prime}\left( x \right) = 0,\;\;}\Rightarrow
    {12\left( {{x^2} – 6x + 8} \right) = 0,\;\;}\Rightarrow
    {{x^2} – 6x + 8 = 0,\;\;}\Rightarrow
    {{x_1} = 2,\;{x_2} = 4.}
    \]

    In this case it is convenient to use the second sufficient condition for the existence of an inflection point. The third derivative is written as

    \[
    {f^{\prime\prime\prime}\left( x \right) }
    = {{\left( {12\left( {{x^2} – 6x + 8} \right)} \right)^\prime } }
    = {12\left( {2x – 6} \right) = 24\left( {x – 3} \right).}
    \]

    From this we immediately see that the third derivative is not zero at the points \({x_1} = 2\) and \({x_2} = 4.\) Therefore, these points are points of inflection.

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    Problems 1-6
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    Problems 7-20