# Is f(x)=cosx concave or convex at x=(3pi)/2?

Feb 9, 2018

See below.

#### Explanation:

We can determine where a function is convex or concave, by using the second derivative. If:

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} f \left(x\right) > 0 \textcolor{w h i t e}{888}$ Convex( concave up )

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} f \left(x\right) < 0 \textcolor{w h i t e}{888}$ Concave( concave down )

The second derivative is just the derivative of the first derivative. .i.e.

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} f \left(x\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \left(\frac{\mathrm{dy}}{\mathrm{dx}} f \left(x\right)\right)$

For $f \left(x\right) = \cos \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \sin \left(x\right)$

$\frac{{d}^{2} y}{{\mathrm{dx}}^{2}} \left(\cos \left(x\right)\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \left(- \sin \left(x\right)\right) = - \cos \left(x\right)$

Now we solve the inequalities:

$- \cos \left(x\right) > 0 \textcolor{w h i t e}{888} \left[1\right]$

$- \cos \left(x\right) < 0 \textcolor{w h i t e}{888} \left[2\right]$

$- \cos \left(x\right) > 0 \textcolor{w h i t e}{888} \left[1\right]$

$\frac{\pi}{2} < x < \frac{3 \pi}{2}$

$- \cos \left(x\right) < 0 \textcolor{w h i t e}{888} \left[2\right]$

$0 < x < \frac{\pi}{2} \textcolor{w h i t e}{88}$ , $\frac{3 \pi}{2} < x < 2 \pi$

Notice that $\boldsymbol{\frac{3 \pi}{2}}$ is on the point where the function changes from convex to concave. This is called a point of inflection ( inflexion in the UK ), so at $\boldsymbol{\frac{3 \pi}{2}}$ it is neither concave nor convex.

This is verified by its graph: