Question

How do I solve for the points of inflection involving trig functions?

`f(x)=sinx+cosx`, `[0, 2π]`

For the derivative I got:
`f'(x)=cosx-sinx`
`f''(x)=-sinx-cosx`
but after that I get lost. Can you please help me?

Answer 1

The points of inflection are at:
`((3pi)/4,0)` and `((7pi)/4,0)`

Explanation:

You were definitely on the right track. In fact, you had two steps remaining. Points of inflection on a graph are where the concavity of the graph changes. In this case, you're looking for the inflection point of:

`f(x)=sinx+cosx` on the interval of `[0, 2pi]`.

The inflection point comes from where the second derivative is equal to 0. So we need to take the derivative twice, which you did perfectly.

`f'(x)=cosx-sinx`
`f''(x)=-sinx-cosx`

Now, it's a matter of seeing where they're equal to one another. So let's solve for x.

`0=-sinx-cosx`
`cosx=-sinx`

Because of the unit circle, we know the value has to be in the second or fourth quadrants, so x will be between `pi/2` and `pi`, and between `3pi/2` and `2pi`. Considering the values are supposed to be equal, we can easily find the solutions, thanks to the unit circle.

https://commons.wikimedia.org

We can see in the image that the functions will be equal at:

`x=(3pi)/4` and `x=(7pi)/4`

So bringing us back to the original question of finding the inflection points, these points are the x values of your inflection points. Now, all you have to do is plug in the values for x into the original function to get your two inflection points. Though considering the second derivative is just the original function multiplied by -1, we can see the points of inflection are at:

`((3pi)/4,0)` and `((7pi)/4,0)`