hey guys can you help me with this:

Use a linear approximation of 3cos^2(3) to estimate the value of |cos(3)|

where do I begin? would it look something like this:

y-(3cos^2(3))=

- May 13th 2011, 07:08 PMRezManLinear Approximation of 3cos^2(3) to estimate the value of |cos(3)
hey guys can you help me with this:

**Use a linear approximation of 3cos^2(3) to estimate the value of |cos(3)|**

where do I begin? would it look something like this:

y-(3cos^2(3))= - May 13th 2011, 08:04 PMTKHunny
Is there a particular linear approximation you had in mind?

The "slope" piece of your line is zero (0). That's not very helpful. - May 14th 2011, 05:15 AMHallsofIvy
Frankly your work doesn't make a whole lot of sense to me. You want are trying to get a linear approximation

**at**x= 3. The slope is, of course, the derivative of which is , evaluated at x= 3 which requires calculating cos(3) and sin(3) anyway.

TKHunny is suggesting you use a linear approximation at rather than at x= 3. There, as TKHunny says, the slope is 0 so your approximation is just the value at which is 3. - May 14th 2011, 06:59 PMRezMan
Well this question is the third part of the initial question, which is:

**Given the function y=xcos^2(x)**which I found to be: cos^2(x)-2xcos(x)sin(x)'

a) find the differential of the function:

**b) Give the linearization of the function at x=pi.**Which I found to be: y=x

and then there's that questionwhich I don't really understand how to do it.

c) Use a linear approximation of 3cos^2(3) to estimate the value of |cos(3)|

I guess I have to find value of x that allow 3cos^2(3) to equal |cos(3)|? - May 14th 2011, 07:15 PMTKHunny
Very good.

a) OK

b) OK

c) Well, what is the approximation? You have A(x) = x, the linearization of y(x) close to pi. x = 3 is pretty close to pi. Use this to create an approximation of y(3). Then use algebra to produce the approximation of |cos(3)|. - May 14th 2011, 08:21 PMRezMan
(Worried) you know I should understand what that means but I don't. Could you please detail it a little more

- May 14th 2011, 08:37 PMTheEmptySet
- May 15th 2011, 07:25 PMRezMan
this is what my teacher says:

*"So in this case, SUPPOSE you had y = 2x - 5 as a linear approximation for y = f(x) near x = 3 (that is not the case, i'm just saying pretend that is your linear approximation).*

then we could say f(3)= 3 cos^2(3) is APPROXIMATELY EQUAL to y = 2(3) - 5 = 1

Since 3 cos^2(3) is APPROXIMATELY EQUAL to 1, we conclude that cos^2(3) is APPROXIMATELY EQUAL to 1/3, so by taking square roots we find |cos(3)| is APPROXIMATELY EQUAL to sqrt (1/3)."

So 3cos^2(3) = 3, because y=x. So maybe cos^2(3) =1? so 3(1)=3. So sqrt(1)=|cos(3)| ? - May 15th 2011, 07:38 PMRezMan
she says yes. Case closed. Thanks guys!