TheCalcHater
  • TheCalcHater
Calc Help? Check my work
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TheCalcHater
  • TheCalcHater
I found the derivative of - x^2-1/(x^2+1)^2. I then set it to zero to get x=1,-1 as our critical points. I then plugged in -2 and 2 into the derivative to see whether it has a negative or positive slope. The relative extrema is x=0 is the local max and x=sqrt3,-sqrt3 is a local minimum.
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zepdrix
  • zepdrix
Your derivative looks a little off, unless you made a typo,\[\large\rm f'(x)=\frac{1-x^2}{(x^2+1)^2}\]the 1 should be positive.
zepdrix
  • zepdrix
Critical points look correct though :) Hmm

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TheCalcHater
  • TheCalcHater
in my calc it shows the whole function to be negative the negative is before the equation...
zepdrix
  • zepdrix
Does it? Ok maybe I made a boo boo :) I'll check again
Astrophysics
  • Astrophysics
That's ok once you distribute you get what zep has
zepdrix
  • zepdrix
Oh did you mean to write this?\[\large\rm f'(x)=-\frac{x^2-1}{(x^2+1)^2}\]
zepdrix
  • zepdrix
I thought the negative was on the x^2 only, my bad.
TheCalcHater
  • TheCalcHater
yes that is what i got
zepdrix
  • zepdrix
I feel like you're missing something for your test points. You wanted to see what was happening "around" each test point. So you chose x=-2 to see what is happening on the left side of x=-1, good. You then chose x=2 to see what is happening on the right side of x=2, good. What about choosing a test point between x=-1 and x=1?
zepdrix
  • zepdrix
to see what happening on the right side of x=1*
TheCalcHater
  • TheCalcHater
ill plug one in give me a sec
TheCalcHater
  • TheCalcHater
if we plug in 0 we get 1 so it is a positive slope
zepdrix
  • zepdrix
|dw:1449864511843:dw|ok sounds good :)
TheCalcHater
  • TheCalcHater
so then my answer is correct i just have to add in that i tested 0?
zepdrix
  • zepdrix
x=0 was not a critical point. How could it possibly be extrema? 0_O
TheCalcHater
  • TheCalcHater
it is the local max
zepdrix
  • zepdrix
??
zepdrix
  • zepdrix
You get your possible max/min values from your critical points. x=0 is not a critical point, so it's not even in the running.
TheCalcHater
  • TheCalcHater
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TheCalcHater
  • TheCalcHater
but when we set it to 0 we get our critical points?
zepdrix
  • zepdrix
This answer key makes no sense at all... where is sqrt(3) coming from? Are you sure this corresponds to this problem?
TheCalcHater
  • TheCalcHater
that is when i plug the derivative into a calc and it says find the local min and max
zepdrix
  • zepdrix
Oh oh oh, this is a different problem. See that f(x) does not match your f(x).
TheCalcHater
  • TheCalcHater
oh should i have used f'(x) or the original f(x)
zepdrix
  • zepdrix
You used f'(x), they wanted f(x) apparently. But I hope what I'm explaining makes sense without you needing the calculator lol :) x=0 was not a critical point, x=-sqrt(3) doesn't even make sense XD lol
Astrophysics
  • Astrophysics
Haha, yeah so essentially what you do is find the derivative and set it equal to 0 and find where it's undefined to find the candidates for your critical values.
Astrophysics
  • Astrophysics
No no, he used f(x) he didn't find f'(x)
zepdrix
  • zepdrix
No, I'm pretty sure his original post includes his f(x), this newer post is some online calculator.
Astrophysics
  • Astrophysics
Haha xD
TheCalcHater
  • TheCalcHater
yeah i found the derivative and set it to 0 to find the critical points... the original f(x) is the other thing the one i posted was the dervative not the original f(x)
TheCalcHater
  • TheCalcHater
ok so where do i take it from here?
Astrophysics
  • Astrophysics
Ok I see you had it listed as f(x)
TheCalcHater
  • TheCalcHater
yeah it should have been f'(x)
zepdrix
  • zepdrix
|dw:1449865034347:dw|you applied the first derivative test, did your sign chart good. Do you see your local max min?
TheCalcHater
  • TheCalcHater
-1 being the min adn 1 being the max?
zepdrix
  • zepdrix
ya :d simple as that :D
TheCalcHater
  • TheCalcHater
thanks i will post the second question in a second
zepdrix
  • zepdrix
Err careful how you say that :) They usually want the y coordinate for the max/min. So we would say x=-1 gives us a minimum value of y=-1/2
TheCalcHater
  • TheCalcHater
ok
TheCalcHater
  • TheCalcHater
here is the second question it uses the MVT
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TheCalcHater
  • TheCalcHater
yeah
zepdrix
  • zepdrix
For us to be allowed to apply the MVT, the function must satisfy two things: ~ Continuous on this interval [0,1] ~ Differentiable on this interval (0,1) Do we have any discontinuities? Does this function have any asymptotes or sharp corners?
TheCalcHater
  • TheCalcHater
Vertical Asymptotes :x=−1 Horizontal Asymptotes :y=0
TheCalcHater
  • TheCalcHater
x=-1 is a discontinuity
zepdrix
  • zepdrix
Discontinuity at x=-1 which is NOT within our interval [0,1]. If we take the derivative of this function, we'll end up with some type of x+1 in the denominator, which still only tells us about x=-1. So good, our function is continuous and differentiable on [0,1], so yes we can apply the MVT :) Step 1 complete.
TheCalcHater
  • TheCalcHater
ok
zepdrix
  • zepdrix
If we take the secant line which connects the ends points of our interval,|dw:1449865638879:dw|
zepdrix
  • zepdrix
|dw:1449865692788:dw|the MVT tells us that there is a tangent line somewhere in this interval with the same slope.
zepdrix
  • zepdrix
(at least one tangent line)
TheCalcHater
  • TheCalcHater
how do we find that tangent line?
zepdrix
  • zepdrix
Remember how to find slope of a line passing through two points?\[\large\rm m=\frac{y_2-y_1}{x_2-x_1}\]Now that we're in Calculus, we try rely heavily on using function notation. So our slope formula instead looks like this,\[\large\rm m=\frac{f(x_2)-f(x_1)}{x_2-x_1}\]
zepdrix
  • zepdrix
Slope of tangent line corresponds to instantaneous rate of change, that's your derivative at a point! :)
TheCalcHater
  • TheCalcHater
so take the derivative? give me a sec.
TheCalcHater
  • TheCalcHater
I got -4(x+1)^-3
zepdrix
  • zepdrix
k good
TheCalcHater
  • TheCalcHater
now what do i do?
zepdrix
  • zepdrix
This derivative corresponds to slope of a tangent line. We'll plug in the value x=c, and get some kind of slope,\[\large\rm m=\frac{-4}{(c+1)^3}\]The MVT tells us that the slope of this tangent line and the slope of the secant line must be equal at some value c.
zepdrix
  • zepdrix
So we'll set them equal, err actually, before we do that, let's calculate our secant line slope first.
TheCalcHater
  • TheCalcHater
so that is the slope of the tangent line. are going to find the equation of the line?
zepdrix
  • zepdrix
No, this is not one of those problems. :)
TheCalcHater
  • TheCalcHater
ok :) so we are going to find the slope of the secant line. is that another derivative?
zepdrix
  • zepdrix
Another derivative? 0_o come on silly billy, get your head in the game!
zepdrix
  • zepdrix
Think back to basic algebra, do you remember finding the slope between two points? I posted the formula a lil while back.
TheCalcHater
  • TheCalcHater
sorry i cant think straight plus i dont know much calc because i skipped pre calc....
TheCalcHater
  • TheCalcHater
slope point formula?
zepdrix
  • zepdrix
Slope is defined to be `change in y` divided by `change in x`. So we'll subtract some y-coordinates, and then divide that by subtracted x-coordinates.
zepdrix
  • zepdrix
\[\large\rm m=\frac{f(x_2)-f(x_1)}{x_2-x_1}\]
TheCalcHater
  • TheCalcHater
ok so what coordinates are we using?
zepdrix
  • zepdrix
For our line, we're using the two points which are the ends of our interval. x=0 and x=1.
zepdrix
  • zepdrix
\[\large\rm m=\frac{f(1)-f(0)}{1-0}\]
TheCalcHater
  • TheCalcHater
so m=1
zepdrix
  • zepdrix
Hmm that doesn't sound right.
zepdrix
  • zepdrix
\[\large\rm f(\color{orangered}{x})=\frac{2}{(\color{orangered}{x}+1)^2}\]So then\[\large\rm f(\color{orangered}{1})=?,\qquad\qquad\qquad f(\color{orangered}{0})=?\]
TheCalcHater
  • TheCalcHater
oh 1/2
TheCalcHater
  • TheCalcHater
and f(0)=2
zepdrix
  • zepdrix
K good, you've got your y-coordinates. Now plug those into your slope formula.
zepdrix
  • zepdrix
The denominator is 1-0=1, so we can ignore that. The numerator is all that matters.
TheCalcHater
  • TheCalcHater
-1.5/1 so -1 1/2
zepdrix
  • zepdrix
Please no mixed numbers -_- ever again... we're not in 3rd grade anymore lol
zepdrix
  • zepdrix
Ok good so -3/2 :)
TheCalcHater
  • TheCalcHater
ok so -3/2 is the slope correct?
zepdrix
  • zepdrix
-3/2 is the slope of the secant line.\[\large\rm m=-\frac{3}{2}\]MTV is telling us that some c value gives us a slope for our tangent line,\[\large\rm m=\frac{-4}{(c+1)^3}\]for which these slopes are equivalent,\[\large\rm -\frac{3}{2}=\frac{-4}{(c+1)^3}\]
zepdrix
  • zepdrix
And the problem is asking you to find the c value(s) that satisfy this. Solve for c.
TheCalcHater
  • TheCalcHater
approx .39
zepdrix
  • zepdrix
Ok good :) And I'm not sure if they want an approximation or exact answer based on the wording of the question.
TheCalcHater
  • TheCalcHater
ok so that is the answer since that satifies c?? thank you soooo much! I will be sure to write a testimonial. Do you happen to know how to balance a net ionic equation? (chemistry)
zepdrix
  • zepdrix
Chemistry? No, fraid not. Sorry if my explanations are long and slow, I know I can take a while to get to the point, that's just how I roll haha :D
TheCalcHater
  • TheCalcHater
no i really enjoy that you explained everything I want to understand calc... I hate people that just give out answers :)

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