This is a calculus problem! If anyone can help me please!!! Urgent!

The problem says: (the pie I don't know how to write the symbol and squared is squaring the sin i don't know how to write it sorry)

Let f be the function defined for pie/6 < x < 5pie/6 by f(x)= x +sin squared x.

it says find:

a.) Find all values of X for which f ' (x)= 1
b.) find the x-coordinates of all minimum points of f. Justify your answer.
c.) find the x-coordinates of all inflection points of f. Justiry your answer.

2. Originally Posted by mmkextreme_1
This is a calculus problem! If anyone can help me please!!! Urgent!

The problem says: (the pie I don't know how to write the symbol and squared is squaring the sin i don't know how to write it sorry)

Let f be the function defined for pie/6 < x < 5pie/6 by f(x)= x +sin squared x.

it says find:

a.) Find all values of X for which f ' (x)= 1
b.) find the x-coordinates of all minimum points of f. Justify your answer.
c.) find the x-coordinates of all inflection points of f. Justiry your answer.

$\displaystyle f(x)= x +sin^2x$

$\displaystyle f'(x) = 1 + 2sin(x)cos(x)$

$\displaystyle 1 = 1 + 2sin(x)cos(x)$

$\displaystyle 2sin(x)cos(x) = 0$

$\displaystyle sin(x)cos(x) = 0$

Can you solve this?

3. Originally Posted by colby2152
$\displaystyle f(x)= x +sin^2x$

$\displaystyle f'(x) = 1 + 2sin(x)cos(x)$

$\displaystyle 1 = 1 + 2sin(x)cos(x)$

$\displaystyle 2sin(x)cos(x) = 0$

$\displaystyle sin(x)cos(x) = 0$

Can you solve this?

no I don't really know what you did there and what would that be the answer to? Part A or what?

4. Originally Posted by colby2152
$\displaystyle f(x)= x +sin^2x$

$\displaystyle f'(x) = 1 + 2sin(x)cos(x)$

$\displaystyle 1 = 1 + 2sin(x)cos(x)$

$\displaystyle 2sin(x)cos(x) = 0$

$\displaystyle sin(x)cos(x) = 0$

Can you solve this?

Actually come to think of it let me see am I suppose to do this:

set sinx=0 and solve to where sin equals 0?

and then set cosx=0 and solve for where cos equals 0?

then where would f'(x)=1?

and then how am I suppose to do b and c of that problem?

I need urgent help on this pleaseee!!!!

5. Originally Posted by mmkextreme_1
Actually come to think of it let me see am I suppose to do this:

set sinx=0 and solve to where sin equals 0?

and then set cosx=0 and solve for where cos equals 0?

then where would f'(x)=1? Mr F says: For the values of x you get after solving cos x = 0 and sin x = 0.
and then how am I suppose to do b and c of that problem?

I need urgent help on this pleaseee!!!!
(b) Solve f'(x) = 0. And to do that, you'll need to:
1. Remember that 2 sin x cos x = sin(2x).
2. Remember how to solve basic trig equations of the form sin(2x) = no.

Then test the nature of the solutions you've found. You want the values of x that give minimum turning points. Then get the corresponding y-coordinates for these values of x.

(c) Solve f''(x) = 0. Then test which of these give points of inflexion..

There are a lot of different mathematical facts and skills you need to apply to successfully answer these questions. These assignment questions have exposed a lot of stuff you need to thoroughly review.

6. Originally Posted by mmkextreme_1
Actually come to think of it let me see am I suppose to do this:

set sinx=0 and solve to where sin equals 0?

and then set cosx=0 and solve for where cos equals 0?
You are right on target...

Originally Posted by mmkextreme_1
then where would f'(x)=1?
Look back at my work, and you will see I do this in the third line.

Originally Posted by mmkextreme_1
and then how am I suppose to do b and c of that problem?

I need urgent help on this pleaseee!!!!
B) Solve for f'(x) = 0
C) Solve for f''(x) = 0

7. mmkextreme: I don't mean to be rude, but these problems are obviously beyond your knowledge base. For example, anyone who doesn't know how to solve $\displaystyle sin(x)~cos(x) = 0$ by the time they are studying Calculus is showing a serious flaw in their basic Math skills.