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Intervals of Concavity

As a continuation of Increasing or Decreasing?, we will now find the intervals of concavity of our function. Whereas the intervals of increasing and decreasing where determined by the first derivative of the function, our intervals of concavity will be determined by the second derivative. That is, our process will be very similar, but we will work with the second derivative instead of the first.

Determine the shape of a function

Recall that we were working with the function
\begin{align*}
f(x)=x(x-2)^{3}.
\end{align*}
With this function we will now want to find

  • intervals of concavity,
  • inflection points.

As we work through this problem, we will work one step at a time.

First Derivative

In order to determine the intervals of concavity, we will first need to find the second derivative of \(f(x)\). Since we found the first derivative in the last post, we will only need to take the derivative of this function. That is, we find that
\begin{align*}
\frac{d^{2}}{dx^{2}}x(x-2)^{3}&=\frac{d}{dx}(x-2)^{2}(4x-2)
\end{align*}
we note that we first have a product rule. Here the first function is \(F=(x-2)^{2}\) and the second is \(S=4x-2\). The derivative of the second is \(F’=4\); however, the derivative of the second requires a chain rule. Here the inside function is \(i=(x-2)\) and the outside function is \(o=i^{2}\). Combining these together, we get
\begin{align*}
\frac{d}{dx}(x-2)^{2}(4x-2)&=SF’+FS’ \\
&=(4x-2)(2(x-2))+(x-2)^{2}*4 \\
&=(8x-4)(x-2)+(4x-8)(x-2) \\
&=(12x-12)(x-2).
\end{align*}

While I could have gone through this in more detail, I will refer you to the post Derivative Examples if you would like more practice with taking derivatives. Further note, that we have taken the time to simplify the derivative because we will be using it.

Finding zeroes

Now that we have the second derivative of \(f(x)\), we will need to determine the points where the second derivative is undefined or \(0\). These points are important because they are the potential inflection points of the function. Furthermore, even if they are not inflection points, these points are still important because they allow us to partition the number line into separate intervals over which the function is either concave up or down, but not both. We then find that
\begin{align*}
f”(x)&=0 \\
(12x-12)(x-2)&=0 \\
(12x-12)&=0 \text{ or } x-2=0 \\
x&=1 \text{ or } x=2.
\end{align*}

Finding Positive and Negative

We can now focus on numbers in the intervals \((-\infty,1)\), \((1,2)\) and \((2,\infty)\). Since the second derivative is not \(0\) or undefined on these intervals, it cannot change from positive to negative. Therefore, if we know what happens at one of the points, we know what happens at all of them. We now find

  • Choose \(x< 1\), say \(x=0\).
  • We want to know if the second derivative is positive or negative, so we find \(f”(0)=24\).
  • Since \(f”(0)=24 >0\), we know \(f”(x) > 0\) for all \(x\) in \((-\infty, 1)\).
  • Choose \(x\) with \(1 < x< 2\), say \(x=1.5\).
  • We want to know if the derivative is positive or negative, so we find \(f”(1.5)=-3\).
  • Since \(f'(1.5)=-3 <0\), we know \(f”(x) < 0\) for all \(x\) in \((1, 2)\).
  • Choose \(x> 2\), say \(x=3\).
  • We want to know if the derivative is positive or negative, so we find \(f”(3)=24\).
  • Since \(f”(3)=24 >0\), we know \(f'(x) > 0\) for all \(x\) in \((2, \infty)\).

We can summarize this with the number line below.

Summarizing

Now that we’ve determined where the second derivative is \(0\), undefined, positive and negative, we can now answer the questions about increasing, decreasing, critical points and extrema.

  • We note that our second derivative is positive on \((-\infty,1) \cup (2,\infty)\).
  • The second derivative is negative on \((1,2)\).
  • Therefore, the function is concave up on \((-\infty,1) \cup (2, \infty)\) and concave down on \((1,2)\).
  • Since the concavity changes at \(x=1\) and \(x=2\) and these are both in the domain, the function has inflection points at both these values.
  • Therefore, the inflection points are \((1,f(1))=(1,-1)\) and \((2,f(2)),(2,0)\).

Conclusion

Here we have continued finding the shape of the graph by determining where the intervals of concavity. As always, I hope this helped with your studies and that you enjoyed the process. Make sure to like this post or share on Social Media if you found it helpful. Also, check back on our next post to see the how the shape can be used to give a sketch of the graph.

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