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?

1 Answer
Jun 14, 2018

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)#