Skip to main content

Subsection 3.6.3 Second Derivative — Concavity

The second derivative \(f''(x)\) tells us the rate at which the derivative changes. Perhaps the easiest way to understand how to interpret the sign of the second derivative is to think about what it implies about the slope of the tangent line to the graph of the function. Consider the following sketches of \(y=1+x^2\) and \(y=-1-x^2\text{.}\)

  • In the case of \(y = f(x) = 1+x^2\) , \(f''(x) = 2 \gt 0\text{.}\) Notice that this means the slope, \(f'(x)\text{,}\) of the line tangent to the graph at \(x\) increases as \(x\) increases. Looking at the figure on the left above, we see that the graph always lies above the tangent lines.
  • For \(y = f(x) = -1-x^2\) , \(f''(x) = -2 \lt 0\text{.}\) The slope, \(f'(x)\text{,}\) of the line tangent to the graph at \(x\) decreases as \(x\) increases. Looking at the figure on the right above, we see that the graph always lies below the tangent lines.

Similarly consider the following sketches of \(y=x^{-1/2}\) and \(y=\sqrt{4-x}\text{:}\)

Both of their derivatives, \(-\frac{1}{2}x^{-3/2}\) and \(-\frac{1}{2}(4-x)^{-1/2}\text{,}\) are negative, so they are decreasing functions. Examining second derivatives shows some differences.

  • For the first function, \(y''(x) = \frac{3}{4}x^{-5/2} \gt 0\text{,}\) so the slopes of tangent lines are increasing with \(x\) and the graph lies above its tangent lines.
  • However, the second function has \(y''(x) = -\frac{1}{4}(4-x)^{-3/2} \lt 0\) so the slopes of the tangent lines are decreasing with \(x\) and the graph lies below its tangent lines.

More generally

Definition 3.6.3

Let \(f(x)\) be a continuous function on the interval \([a,b]\) and suppose its first and second derivatives exist on that interval.

  • If \(f''(x) \gt 0\) for all \(a \lt x \lt b\text{,}\) then the graph of \(f\) lies above its tangent lines for \(a \lt x \lt b\) and it is said to be concave up.

  • If \(f''(x) \lt 0\) for all \(a \lt x \lt b\text{,}\) then the graph of \(f\) lies below its tangent lines for \(a \lt x \lt b\) and it is said to be concave down.

  • If \(f''(c)=0\) for some \(a \lt c \lt b\text{,}\) and the concavity of \(f\) changes across \(x=c\text{,}\) then we call \((c,f(c))\) an inflection point.

Note that one might also see the terms

  • “convex” or “convex up” used in place of “concave up”, and
  • “concave” or “convex down” used to mean “concave down”.

To avoid confusion we recommend the reader stick with the terms “concave up” and “concave down”.

Let's now continue Example 3.6.2 by discussing the concavity of the curve.

Consider again the function

\begin{align*} f(x) &= x^4-6x^3 \end{align*}
  • Its first derivative is \(f'(x)=4x^3-18x^2\text{,}\) so
    \begin{align*} f''(x) &= 12x^2 - 36x = 12x(x-3) \end{align*}

  • Thus the second derivative is zero (and potentially changes sign) at \(x=0,3\text{.}\) Thus we should consider the sign of the second derivative on the following intervals

    \begin{align*} (-\infty,0) && (0,3) && (3,\infty) \end{align*}

    A little algebra gives us

    interval \((-\infty,0)\) 0 \((0,3)\) 3 \((3,\infty)\)
    \(f''(x)\) positive 0 negative 0 positive
    concavity up inflection down inflection up

    Since the concavity changes at both \(x=0\) and \(x=3\text{,}\) the following are inflection points

    \begin{align*} (0,0) && (3,3^4-6\times3^3)=(3,-3^4) \end{align*}
  • Putting this together with the information we obtained earlier gives us the following sketch

In our Definition 3.6.3, concerning concavity and inflection points, we considered only functions having first and second derivatives on the entire interval of interest. In this example, we will consider the functions

\begin{equation*} f(x)=x^{1/3}\qquad g(x)=x^{2/3} \end{equation*}

We shall see that \(x=0\) is a singular point for both of those functions. There is no universal agreement as to precisely when a singular point should also be called an inflection point. We choose to extend our definition of inflection point in Definition 3.6.3 as follows. If

  • the function \(f(x)\) is defined and continuous on an interval \(a\lt x\lt b\) and if
  • the first and second derivatives \(f'(x)\) and \(f''(x)\) exist on \(a\lt x\lt b\) except possibly at the single point \(a\lt c\lt b\) and if
  • \(f\) is concave up on one side of \(c\) and is concave down on the other side of \(c\)

then we say that \(\big(c\,,\,f(c)\big)\) is an inflection point of \(y=f(x)\text{.}\) Now let's check out \(y=f(x)\) and \(y=g(x)\) from this point of view.

  1. Features of \(y=f(x)\) and \(y=g(x)\) that are read off of \(f(x)\) and \(g(x)\text{:}\)

    • Since \(f(0)=0^{1/3}=0\) and \(g(0)=0^{2/3}=0\text{,}\) the origin \((0,0)\) lies on both \(y=f(x)\) and \(y=g(x)\text{.}\)
    • For example, \(1^3=1\) and \((-1)^3=-1\) so that the cube root of \(1\) is \(1^{1/3}=1\) and the cube root of \(-1\) is \((-1)^{1/3}=-1\text{.}\) In general,
      \begin{equation*} x^{1/3}\begin{cases} \lt 0 \amp \text{if }x\lt 0 \\ =0 \amp \text{if }x=0 \\ \gt 0 \amp \text{if }x> 0 \end{cases} \end{equation*}
      Consequently the graph \(y=f(x)=x^{1/3}\) lies below the \(x\)-axis when \(x\lt 0\) and lies above the \(x\)-axis when \(x>0\text{.}\) On the other hand, the graph \(y=g(x)=x^{2/3}=\big[x^{1/3}\big]^2\) lies on or above the \(x\)-axis for all \(x\text{.}\)
    • As \(x\rightarrow+\infty\text{,}\) both \(y=f(x)=x^{1/3}\) and \(y=g(x)=x^{2/3}\) tend to \(+\infty\text{.}\)
    • As \(x\rightarrow-\infty\text{,}\) \(y=f(x)=x^{1/3}\) tends to \(-\infty\) and \(y=g(x)=x^{2/3}\) tends to \(+\infty\text{.}\)
  2. Features of \(y=f(x)\) and \(y=g(x)\) that are read off of \(f'(x)\) and \(g'(x)\text{:}\)
    \begin{align*} f'(x) \amp= \left.\begin{cases} \tfrac{1}{3}x^{-2/3} \amp \text{if }x\ne 0 \\ \text{undefined} \amp \text{if }x =0 \end{cases}\right\} \implies f'(x)\gt 0\text{ for all }x\ne 0 \\ g'(x) \amp= \left.\begin{cases} \tfrac{2}{3}x^{-1/3} \amp \text{if }x\ne 0 \\ \text{undefined} \amp \text{if }x =0 \end{cases}\right\} \implies g'(x) \begin{cases}\lt 0\amp\text{if }x\lt 0\\ \gt 0\amp\text{if }x\gt 0 \end{cases} \end{align*}
    So the graph \(y=f(x)\) is increasing on both sides of the singular point \(x=0\text{,}\) while the graph \(y=g(x)\) is decreasing to the left of \(x=0\) and is increasing to the right of \(x=0\text{.}\) As \(x\rightarrow 0\text{,}\) \(f'(x)\) and \(g'(x)\) become infinite. That is, the slopes of the tangent lines at \(\big(x,f(x)\big)\) and \(\big(x,g(x)\big)\) become infinite and the tangent lines become vertical.
  3. Features of \(y=f(x)\) and \(y=g(x)\) that are read off of \(f''(x)\) and \(g''(x)\text{:}\)
    \begin{align*} f''(x) \amp= \left.\begin{cases} -\tfrac{2}{9}x^{-5/3} =-\tfrac{2}{9}{\big[x^{-1/3}]}^{5} \amp \text{if }x\ne 0 \\ \text{undefined} \amp \text{if }x =0 \end{cases}\right\} \\ \amp \implies f''(x) \begin{cases}\gt 0 \amp \text{if }x\lt 0\\ \lt 0 \amp \text{if }x\gt 0 \end{cases} \\ g''(x) \amp= \left.\begin{cases} -\tfrac{2}{9}x^{-4/3} =-\tfrac{2}{9}{\big[x^{-1/3}]}^{4} \amp \text{if }x\ne 0 \\ \text{undefined} \amp \text{if }x =0 \end{cases}\right\} \\ \amp \implies g''(x)\lt 0\text{ for all }x\ne 0 \end{align*}
    So the graph \(y=g(x)\) is concave down on both sides of the singular point \(x=0\text{,}\) while the graph \(y=f(x)\) is concave up to the left of \(x=0\) and is concave down to the right of \(x=0\text{.}\)

By way of summary, we have, for \(f(x)\text{,}\)

\((-\infty,0)\) 0 \((0,\infty)\)
\(f(x)\) negative 0 positive
\(f'(x)\) positive undefined positive
increasing increasing
\(f''(x)\) positive undefined negative
concave up inflection concave down

and for \(g(x)\text{,}\)

\((-\infty,0)\) 0 \((0,\infty)\)
\(g(x)\) positive 0 positive
\(g'(x)\) negative undefined positive
decreasing increasing
\(g''(x)\) negative undefined negative
concave down inflection concave down

Since the concavity changes at \(x=0\) for \(y=f(x)\text{,}\) but not for \(y=g(x)\text{,}\) \((0,0)\) is an inflection point for \(y=f(x)\text{,}\) but not for \(y=g(x)\text{.}\) We have the following sketch for \(y=f(x)=x^{1/3}\)

and the following sketch for \(y=g(x)=x^{2/3}\text{.}\)

Note that the curve \(y=f(x)=x^{1/3}\) looks perfectly smooth, even though \(f'(x)\rightarrow\infty\) as \(x\rightarrow 0\text{.}\) There is no kink or discontinuity at \((0,0)\text{.}\) The singularity at \(x=0\) has caused the \(y\)-axis to be a vertical tangent to the curve, but has not prevented the curve from looking smooth.

login