how many points of inflection are there for the function y=x+cos(2x) on the interval [0,pi]

Question

solomonzelman · Answer

Note:
There is a test that determines whether or not your inflection point is actually an inflection or not, but let's leave that for later.

solomonzelman · Answer

Do you have any problems with finding the second derivative?

solomonzelman · Answer

(( just making sure, is it calculus, or is it not? ))

lillian_a · Answer

yes it's calculus

lillian_a · Answer

its a calculator problem

solomonzelman · Answer

ti-84 calculator problem?

lillian_a · Answer

yes

lillian_a · Answer

second derivative: -4cos(2x)

solomonzelman · Answer

Oh, that was very good!

solomonzelman · Answer

$\color{#000000 }{ \displaystyle f''(x)=-4\cos(2x)  }$

solomonzelman · Answer

We have a constraint: $\color{#000000 }{ \displaystyle x \in [0,\pi]  }$
(i.e.   $\color{#000000 }{ \displaystyle 0 \le x \le\pi }$)

So, set,  $\color{#000000 }{ \displaystyle 0=-4\cos(2x)  }$
And tell me which x-solutions (if any) are within this interval

solomonzelman · Answer

(I don't recall any ti-calculator-functions that relate to finding infleection points, although we can do it with "normal" calculus)

lillian_a · Answer

how do i figure out the root between 0 and pi

solomonzelman · Answer

for what values of x is -4cos(2x)=0 ?

solomonzelman · Answer

can you list a couple of first positive values for which cos(x)=0?

lillian_a · Answer

Im not sure