How Can We Determine the Max Min of Trig Functions in the Range (0, pi)?

[Somefantastik]

x in (0,pi];

f(x) = sin(x)-x²;

f'(x) = cos(x) - 2x;

f'(x) = 0 ==> cos(x) - 2x = 0;

since |cos(x)| ≤ 1,

cos(x) - 2x ≤ 1 - 2x;

Now 1-2x = 0 <==> x = 1/2;

f'(1/4) = cos(1/4) - 2*(1/4) > 0 and f'(3/4) = cos(3/4) - 2*(3/4) < 0;

==> x = 1/2 is maximum and f'(x) ≤ 1/2;

Is my logic correct?

 

[HallsofIvy]

x in (0,pi];

f(x) = sin(x)-x²;

f'(x) = cos(x) - 2x;

f'(x) = 0 ==> cos(x) - 2x = 0;

since |cos(x)| ≤ 1,

cos(x) - 2x ≤ 1 - 2x;

Now 1-2x = 0 <==> x = 1/2;

The fact that 1- 2x= 0 does NOT mean cos(x)- 2x= 0 since cos(x)- 2x< 1- 2x, if 1- 2x= 0 then cos(x)- 2x< 0. In fact, with x= 1/2, cos(1/2)- 2(1/2)= -1.22. cos(x)- 2x= 0 when cos(x)= 2x which happens when x is approximately .45.

f'(1/4) = cos(1/4) - 2*(1/4) > 0 and f'(3/4) = cos(3/4) - 2*(3/4) < 0;

==> x = 1/2 is maximum and f'(x) ≤ 1/2;

Is my logic correct?

No, it is not. saying that cos(x)- 2x< 1- 2x= 0 does not mean cos(x)- 2x= 0.

 

[Somefantastik]

I'm trying to find the max and min of this function. Can someone give me some advice?

 

[Mark44]

You'll have to do this by numerical approximation, since the equation is not solvable by analytic means.

cos (x) - 2x = 0 <==> x = .5 cos(x)

Start with an initial x value of your choice.
Evaluate .5*cos(x) at that value to get your next x value.
Repeat step 2.
When the input value and output value are "close enough" that's your approximate solution.

 

[Somefantastik]

How about just getting a bound on this function ( f'(x) ) on (0,pi]; can I do that without finding the max and min?

f'(0) = 1 and f'(pi) = -1 - 2*pi

is my guess, but I am unsure of any bigger fluctuations on that interval.

 

[Studiot]

How about just getting a bound on this function ( f'(x) ) on (0,pi]; can I do that without finding the max and min?

Well let's look at the function shall we?

sinx is always less than 1.
the max slope of sinx occurs at zero and equals 1; thereafter it is always below the line y=x.
at small values

\sin x\quad \simeq x

since x \le {(x)^2}

and - {(x)^2} decreases monotonically from 0 to - {(\pi )^2} \simeq 10

The function is negative once x \le {(x)^2} is greaer than 1.

Can you sketch it?