TheCalcHater
  • TheCalcHater
I need some calculus help....
Mathematics
  • Stacey Warren - Expert brainly.com
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
What's your question?
IrishBoy123
  • IrishBoy123
\[ f(x)=\dfrac{x^2}{x^2+81} = 1- 81(x^2+81)^{-1} \] \(f' = 81(2x)(x^2+81)^{-2}\) is that where you are because it looks like you are correct?
TheCalcHater
  • TheCalcHater
Find the value of derivative (if it exists) of the function f(x)7-|x| at the extremum point (0,7)

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TheCalcHater
  • TheCalcHater
0 Does not Exist -8 8
anonymous
  • anonymous
what is f(x)?
anonymous
  • anonymous
oh okay
anonymous
  • anonymous
so it is f(x)= 7- abs{x} ?
IrishBoy123
  • IrishBoy123
\((|x|)'\) is not defined
anonymous
  • anonymous
the derivative is abs{x}/x
anonymous
  • anonymous
sorry I forgot the minus
anonymous
  • anonymous
it does exist
TheCalcHater
  • TheCalcHater
can you put the steps and answer and walk me through it
anonymous
  • anonymous
abs{x}= x for x>=0 and -x for x=<0
TheCalcHater
  • TheCalcHater
so 0?
TheCalcHater
  • TheCalcHater
it only count as participation, but I want to understand it
amistre64
  • amistre64
what does it mean to 'be a limit'?
amistre64
  • amistre64
hmm, well good try at defining it. i was leaning more towards ... a limit is defined if and only if all paths lead to the same thing as we approach it.
amistre64
  • amistre64
now for your case, a derivative is a limit ... the limit of a slope. is the slope from the left and the right the same as we approach 0,7?
TheCalcHater
  • TheCalcHater
yes?
TheCalcHater
  • TheCalcHater
(will you be able to help me with a few more after this one?)
amistre64
  • amistre64
graphing it may be useful .. |dw:1449791491297:dw|
TheCalcHater
  • TheCalcHater
so no...
amistre64
  • amistre64
at the point 0,7. the slope from the left is 1, the slope from the right is -1. since the value is not the same for all directions that we can approach from .. we say that the limit is not defined .. does not exist.
TheCalcHater
  • TheCalcHater
ok thanks ill post the second one give me a minute
TheCalcHater
  • TheCalcHater
locate the absolute extrema of the function f(x)=cos(pix) on the closed interval [0,3/4]
amistre64
  • amistre64
what should we test for?
TheCalcHater
  • TheCalcHater
if it is continous on the interval?
amistre64
  • amistre64
cos is a continuous function ... what else, regarding min and max do you think would be useful?
TheCalcHater
  • TheCalcHater
trying to find the mix and max on that interval?
amistre64
  • amistre64
yes, using the derivative set equal to zero we can determine some min/max critical points ... and also testing the end points.
TheCalcHater
  • TheCalcHater
ok so?
amistre64
  • amistre64
so what is our derivative of cos(pi x)
TheCalcHater
  • TheCalcHater
-pisin(pix)
amistre64
  • amistre64
and when that is set to equal zero, what do we get for our solution set?
TheCalcHater
  • TheCalcHater
my calc says +-2n, pi +-2n
TheCalcHater
  • TheCalcHater
that is where i am stuck
amistre64
  • amistre64
any integer value of x will make sin(pix)=0
TheCalcHater
  • TheCalcHater
so is it no max and min f(0)=1?
amistre64
  • amistre64
|dw:1449792069863:dw|
amistre64
  • amistre64
graphically, we have a max at x=0, and the other extrema on the interval is well, at the other endpoint in this case. x=3/4
TheCalcHater
  • TheCalcHater
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TheCalcHater
  • TheCalcHater
so b?
amistre64
  • amistre64
yes, b seems fine to me
TheCalcHater
  • TheCalcHater
Ok thanks here is the next one
1 Attachment
TheCalcHater
  • TheCalcHater
(after this question i will close it and start a new one) ok? or do u want me to just keep going on this one?
amistre64
  • amistre64
define the thrm and its required conditions
TheCalcHater
  • TheCalcHater
Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b), then there is at least one point c in (a, b) where f '(c) = 0.
amistre64
  • amistre64
we are continuous .. its a quadratic polynomial ... what is the value of f(a) and f(b)?
TheCalcHater
  • TheCalcHater
(-4,-2)??
amistre64
  • amistre64
not quite .. what is the value of the function when x=-4? and when x=-2?
TheCalcHater
  • TheCalcHater
0 and -8
amistre64
  • amistre64
does 0 = -8 then?
TheCalcHater
  • TheCalcHater
no
TheCalcHater
  • TheCalcHater
so it does not apply?
amistre64
  • amistre64
so, since f(a) not equal f(b) i would say it does not apply. that, we have a slope of 0 when x=-1 ... which isnt in the interval anyways
TheCalcHater
  • TheCalcHater
so it doesnt apply?
amistre64
  • amistre64
id go with that yes
TheCalcHater
  • TheCalcHater
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amistre64
  • amistre64
and your thoughts?
TheCalcHater
  • TheCalcHater
I want to guess b but i have no clue...
amistre64
  • amistre64
no need to guess. start by defining the thrm in the problem
TheCalcHater
  • TheCalcHater
It states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that
amistre64
  • amistre64
also, we would want a number between 0 and 10 ... so anything negative would not be in that interval by default
amistre64
  • amistre64
so, define the slope between the endpoints
TheCalcHater
  • TheCalcHater
I'm lost
amistre64
  • amistre64
your doing calculus ... which means that you have already passed the course that requires you to find the slope between 2 points ...
TheCalcHater
  • TheCalcHater
what two points? Also I never had pre calc
amistre64
  • amistre64
pre calc is trig ... the slope between 2 points is algebra
TheCalcHater
  • TheCalcHater
oh ok lets continue
amistre64
  • amistre64
you have a function defined, you have an interval defined. use the function, and the endpoints of the interval to determine the slope
TheCalcHater
  • TheCalcHater
you mean the two end pointe 0 and 10?
amistre64
  • amistre64
yes, and the function x^3
TheCalcHater
  • TheCalcHater
if so that would be a vertical slope so undefined
TheCalcHater
  • TheCalcHater
I'm confused
amistre64
  • amistre64
its not vertical ... in fact the problem you posted already tells you how to define it. f(b)-f(a) ------- b-a
amistre64
  • amistre64
10^3 - 0^3 ---------- = 100 10 - 0
TheCalcHater
  • TheCalcHater
oh i see now
TheCalcHater
  • TheCalcHater
then what?
amistre64
  • amistre64
we need to find a point that has a slope of 100 ... how do we find the slope at a point?
TheCalcHater
  • TheCalcHater
take the derivative?
amistre64
  • amistre64
yes and ive read the question over again ... and something is off. it wasnt you to find a slope using x=16 which is NOT part of the interval.
TheCalcHater
  • TheCalcHater
so the answer would be D?
amistre64
  • amistre64
not sure 16^2 is what 256? 3x^2 = 256 16/sqrt(3) is right ... but it does not really help with the setup of the question. the mean value thrm is for finding the slope between the endpoints, and finding a value in the interval that matches that slope. they have asked for a slope that is outside of the interval .... (0,10) does not contain 16.
TheCalcHater
  • TheCalcHater
if we cant figure that one out here is the next one
1 Attachment
TheCalcHater
  • TheCalcHater
this i have absolutely no clue on how to do
amistre64
  • amistre64
it might as well have asked you to find the value of c such that f'(c)=-3 for all that matters.
TheCalcHater
  • TheCalcHater
did you get the next one?
amistre64
  • amistre64
its oddly worded ... since nothing is 'given' i assume it means that you need to find the information instead.
TheCalcHater
  • TheCalcHater
do you want another problem instead adn ill guess on that one?
amistre64
  • amistre64
no need to guess, im just saying its poorly written that, and i have to be on my way
amistre64
  • amistre64
what is the derivative of f?
TheCalcHater
  • TheCalcHater
awww i had 5 more... Thanks for your help though :)
TheCalcHater
  • TheCalcHater
give me a sec
TheCalcHater
  • TheCalcHater
6(x-1)^-7/13 all divided by 13
amistre64
  • amistre64
since our exponent is negative ... we have a critical value when x-1 = 0
TheCalcHater
  • TheCalcHater
ok then what?
amistre64
  • amistre64
when does x-1=0?
TheCalcHater
  • TheCalcHater
when x=1?
amistre64
  • amistre64
yes, so our only real options are the first 2, and they differ in one aspect ... if you start by going uphill (increasing) and then go downhill (decreasing) then would we have been at the highest point, or the lowest point?
TheCalcHater
  • TheCalcHater
highest right becasue at the op of the hill it would be the max and the vallen (down) would be the min
amistre64
  • amistre64
highest, max correct :) and suppers on so good luck with the rest
TheCalcHater
  • TheCalcHater
so it would be a or b?
TheCalcHater
  • TheCalcHater
a correct?
TheCalcHater
  • TheCalcHater
for those of you that are viewing if i close this and open another can you help me?
TheCalcHater
  • TheCalcHater
??

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