One of the many interesting things I’ve come across while learning calculus is the concept of concavity. A simple example of a concave function is a parabola. The left graph below shows a concave up function: \(\boldsymbol{y = x^{2}}\), while the right graph shows a concave down function: \(\boldsymbol{y = -x^{2}}\).
I think this was pretty obvious looking at the graphs. But, if we have a more complicated function, how do we find concavity analytically, without looking at a graph? A function can be concave up within an interval and then change to being concave down in another interval. How can we tell where a function is concave up or down and where an Inflection Point exists; that is, where does concavity change?
I’m going to use as an example the function: \(\normalsize{f\!\left(x\right) = x^{4}\! -\, 4x^{3} + 7x + 11}\). This is a continuous function in its entire domain \(\boldsymbol{\left(-\infty , \infty\right)}\).
How do we proceed? As Prof. Edwards, co-author of many calculus textbooks and lecturer in several of The Great Courses video series, says: “When in doubt, take the derivative.”
\begin{align*}
\text{Let’s do that}&: & \frac{d}{dx} \left(x^{4}\! -\, 4x^{3} + 7x + 11\right) &= 4x^{3}\! -\, 12x^{2} + 7 \\
\text{Set the derivative to zero}&: & 4x^{3}\! -\, 12x^{2} + 7 &= 0 \\
\text{and solve for x}&: & x &= -0.68878,\, 0.91647,\, 2.77230
\end{align*}
We can do a First Derivative Test to find out whether, at these values of \(\boldsymbol{x}\), the function \(f(x)\) has local maxima or minima. But we are interested in concavity and inflection points. For this, we need to take the Second Derivative, that is, the derivative of the first derivative.
\begin{align*}
\text{Let’s do that}&: \frac{d}{dx} \left(4x^{3}\! -\, 12x^{2} + 7\right) = 12x^{2}\! -\, 24x \\
\text{Next, evaluate the 2nd derivative}&: 12x^{2}\! -\, 24x \big|_{x=-0.68878} > 0 \; \text{(concave up)} \\
\text{Next, evaluate the 2nd derivative}&: 12x^{2}\! -\, 24x \big|_{x=0.91647} < 0 \, \text{(concave down)} \\
\text{Next, evaluate the 2nd derivative}&: 12x^{2}\! -\, 24x \big|_{x=2.77230} > 0 \, \text{(concave up)} \\
\text{Set the 2nd derivative to zero}&: 12x^{2}\! -\, 24x\, = 0 \\
\text{and solve for x}&: \qquad \quad \qquad = 0, \, 2 \\
& f(0) = 11, \, f(2) = 9
\end{align*}
Looking at the results above, we conclude that the function \(\normalsize{f(x) = x^{4}\! -\, 4x^{3} + 7x + 11}\) is:
\begin{align*}
\text{Concave up on: } & \left(-\infty,0\right) \\
\text{Concave down on: } & \left(0,2\right) \\
\text{Concave up on: } & \left(2,\infty\right) \\
\text{Inflection points are: } & \left(0,11\right), \, \left(2,9\right)
\end{align*}
The moment of truth. Let’s see if the graph of \(\normalsize{f\!\left(x\right) = x^{4}\! -\, 4x^{3} + 7x + 11}\) agrees with our conclusions.
In summary, there are several tools available using calculus to analyze functions.
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .