# Thread: Graph-min, inflection pts for a trig function

1. ## Graph-min, inflection pts for a trig function

Usually I'm fine with most of the graphing..but the trig always trips me up..
The problem:

Let f be the function defined for x is greater than or equal to pi/6 and x is less than or equal to 5pi/6. F(x)= x+sin^2(x)

a) Find all values of x for which f(x)=1

I did it on my calculator and got x is approximately 0.6417...by hand I tried to factor w/ 0=x+sin^2-1 but that was a fail...So is my answer exact enough, or how would I get the exact answer by hand?

b) Find the x-coordinates for all min points of f.

So I take the f'(x), which = 1+2sin(x)cos(x), then to simply that I got it down to -1=sin(2x). What do I do from here?

c) find the inflection pts...so I just take the second derivative right?

2. Originally Posted by bcahmel
Usually I'm fine with most of the graphing..but the trig always trips me up..
The problem:

Let f be the function defined for x is greater than or equal to pi/6 and x is less than or equal to 5pi/6. F(x)= x+sin^2(x)

a) Find all values of x for which f(x)=1

I did it on my calculator and got x is approximately 0.6417...by hand I tried to factor w/ 0=x+sin^2-1 but that was a fail...So is my answer exact enough, or how would I get the exact answer by hand?
$\displaystyle sin^2x=1-cos^2x\Rightarrow\ x+sin^2x-1=0\Rightarrow\ x-cos^2x=0\Rightarrow\ (\sqrt{x}-cosx)(\sqrt{x}+cosx)=0$

You could try the Newton-Raphson method to solve

$\displaystyle cos^2x-x=0$

but you will get a very close approximation.

3. ok thanks for help with that part...good to know I didn't learn that method in trig!