Can a function have curvature if it has no inflection point

Question

The function $$f(x) = \dfrac{1}{x^2-1}$$

has no inflection point, but can it have curvature (concave/convex) ?

Answer 1

Even though this function has no inflection point, it has infinite discontinuities and its concavity can change on either side of that discontinuity. In this function, the concavities are as follows:

• For $x < -1$, the derivative is increasing, so the function is concave up.
• For $-1 < x < 1$, the derivative is decreasing, so the function is concave down.
• For $x > 1$, the derivative is increasing, so the function is concave up.

This is clear from the graph, but we can also figure this out using the second derivative. First, rewrite the function using partial fraction decomposition:

$$f(x)=\frac{0.5}{x-1}-\frac{0.5}{x+1}=0.5(x-1)^{-1}-0.5(x+1)^{-1}$$

Then, use power rule to find the second derivative:

$$f'(x)=0.5*-1(x-1)^{-2}-0.5*-1(x+1)^{-2}=0.5(x+1)^{-2}-0.5(x-1)^{-2}$$ $$f''(x)=0.5*-2(x+1)^{-3}-0.5*-2(x-1)^{-3}=(x-1)^{-3}-(x+1)^{-3}=\frac 1 {(x-1)^3}-\frac 1 {(x+1)^3}$$

Then, add the fractions so you have a common denominator:

$$f''(x)=\frac 1 {(x-1)^3}-\frac 1 {(x+1)^3}=\frac{(x+1)^3-(x-1)^3}{(x+1)^3*(x-1)^3}=\frac{6x^2+2}{(x^2-1)^3}$$

Notice how the numerator is always positive because it is a square (always positive) times a positive number plus a positive number. The denominator is $0$ at $x=-1$ and $x=1$, so we know we need to break this up into the following cases:

• $x < -1$ -> Here, $\lvert x \rvert > 1$, meaning $x^2 > 1$, meaning $x^2-1 > 0$, meaning $(x^2-1)^3 > 0$, so the denominator is positive. This, combined with the fact that the numerator is positive, means that the whole second derivative is positive, so the function is concave up.
• $-1 < x < 1$ -> Here, $\lvert x \rvert < 1$, meaning $x^2 < 1$, meaning $x^2-1 < 0$, meaning $(x^2-1)^3 < 0$, so the denominator is negative. This, combined with the fact that the numerator is positive, means that the whole second derivative is negative, so the function is concave down.
• $x > 1$ -> Here, $\lvert x \rvert > 1$, meaning $x^2 > 1$, meaning $x^2-1 > 0$, meaning $(x^2-1)^3 > 0$, so the denominator is positive. This, combined with the fact that the numerator is positive, means that the whole second derivative is positive, so the function is concave up.

Thus, using the second derivative, we were able to find the concavity of this function even though it doesn't have inflection points.