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Lim as x approaches pi/6 of (sin(x)-sin(pi/6))/x-pi/6

Mathematics
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compare to definition of derivative
\[\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(a)\]
So the limit would = 1?

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Other answers:

no
what is f?
...?
what is f(x) and what is a?
F(x)=sin(x), f(a)=sin(pi/6)
so what is f'?
Cos(x)-cos(pi/6)?
:( if f=sin(x), then f'=?
Cos(x)
right now plug in a to get your answer
A=sin(pi/6)?
\[\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}=f'(a) \text{ or we could say it equals } =(f(x))'|_{x=a}\]
\[\lim_{x \rightarrow \frac{\pi}{6}}\frac{\sin(x)-\sin(\frac{\pi}{6})}{x-\frac{\pi}{6}}=(\sin(x))'|_{x=\frac{\pi}{6}} \]
do you see all i'm trying to get you to do is use the definition of derivative
you are suppose to determine f and a then find f' then plug a into f' to get the result you are searching for
Sqrt(3)/2?
yep
Thank you
Any chance you can help with critical numbers?
sure
F(x)=4x/x^2+1
I know i need the derivative
do you mean 4x/(x^2+1)? Also what did you get for the derivative?
Yes. F'(x)=(-4x^2+4)/(x^2+1)^2
thats great now find when F'=0 Also we don't have to worry when F' DNE because x^2+1 is always positive. so F'=0 when top=0 so when -4x^2+4=0 solve for x now
Sqrt(-1)
I think you should retry solving -4x^2+4=0 I will give you a (you will get two real solutions)
...+or-sqrt(-1)
+or-sqrt(1)
right sqrt(1)=1 so x=1 or x=-1
those are the critical numbers for F.
Wow...
Find the intervals where f(x)=4/(x^+1) is increasing or decreasing.
is that f(x)=4/(x+1)? like you weren't trying to write an exponent right?
I Find the derivative...then set the denominator equal to zero...put those numbers on a number line and plug in numbers to find the intervals?
I was writing an exponent
I apologize for missing the exponent
It should have said 4/(x^2+1)
well I don't know what f you are looking at exactly. but yeah you definitely need to find f' because f' tells you if f is decreasing or increasing or neither f'=0 means the function is resting f'>0 means f is increasing f'<0 means f is decreasing I first find when f' =0 and when f' does not exist and then test each interval around those numbers to see if the function is increasing there or decreasing there
so what is f' first?
-8x/(x^4+1)
X=-1 and x=1
(x^2+1)^2 does not equal x^4+1
just leave is at (x^2+1)^2
Increasing from negative infinity to 0 and decreasing from 0 to positive infinity
also x^2+1 is always positive so f' and f exist everywhere so you just need to find when f'=0 which is when -8x=0 which is when x=?
ok yep
F(x)=3x^2-12x=5 over the closed interval 0,3. Find absolute extrema.
Do i again begin with taking the derivative?
see if there any critical numbers between 0 and 3 first
which means you need to find f' which all means you need find when f'=0 this looks like you meant to write a polynomial so we don't need to worry when f' dne
X=2
ok do F(0) and F(2) and F(3) which ever one of these are the lowest value is the absolute min value which ever one of these are the highest value is the absolute max value
Ok so you use the tow numbers in the interval and the critical number?
also I don't know what your function is exactly critical numbers in between the critical numbers and any included endpoints just plug them back into the original function
you have two equal signs
i'm guessing that second equal sign is a + or a -
can't decide which one
Should be +5
ok do you want me to check your answers for this one or are you just going through problems and getting ideas on how to do them so you can pass your final
like it seems like you are studying
I just need ideas on how to pass my final...i struggle with a few areas.
i would like to help you more but...
I have to go do some house stuff good luck on your final
i hope i was helpful
Ok thanks

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