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))=$\displaystyle \frac{d(3cos^2(3))}{dy}(x-(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))=$\displaystyle \frac{d(3cos^2(3))}{dy}(x-(3)) $
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 $\displaystyle 3cos^2(x)$ which is $\displaystyle -6cos(x)sin(x)$, evaluated at x= 3 which requires calculating cos(3) and sin(3) anyway.
TKHunny is suggesting you use a linear approximation at $\displaystyle x= \pi$ rather than at x= 3. There, as TKHunny says, the slope is 0 so your approximation is just the value at $\displaystyle x= \pi$ which is 3.
Well this question is the third part of the initial question, which is:
Given the function y=xcos^2(x)
a) find the differential of the function: which I found to be: cos^2(x)-2xcos(x)sin(x)'
b) Give the linearization of the function at x=pi. Which I found to be: y=x
and then there's that question
c) Use a linear approximation of 3cos^2(3) to estimate the value of |cos(3)|
which I don't really understand how to do it.
I guess I have to find value of x that allow 3cos^2(3) to equal |cos(3)|?
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)| ?