Question

Use Theorem 11.24 to prove that the curvature is zero at a point of inflection of a twice-differentiable function y = f(x).

Short answer

It is proved thatthe curvature is zero at a point of inflection of a twice-differentiable functiony = f(x).

Step-by-step solution

Step 1. Given Information. 

It is given that thetwice-differentiable function isy = f(x).

Step 2. Prove. 

To prove that the curvature is zero at a point of inflection of a twice-differentiable function y = f(x),we will use the formula for Curvature in the Plane.

The curvature in the plane is defined as letting y = f(x) be a twice-differentiable function. Then the curvature of the graph of f is given by $k=\frac{\left|f\text{'}\text{'}\left(x\right)\right|}{{\left(1+{\left(f\text{'}\left(x\right)\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}.$

Now, let xbe a point of inflection of a twice-differentiable function f.Thus, at each inflection point, $f\text{'}\text{'}\left(x\right)=0.$

So,

$$\begin{gathered} k=\frac{\left|f\text{'}\text{'}\left(x\right)\right|}{{\left(1+{\left(f\text{'}\left(x\right)\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ k=\frac{0}{{\left(1+{\left(f\text{'}\left(x\right)\right)}^2\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}} \\ k=0 \end{gathered}$$

Hence proved.