Help finding the maxima and minima

-Whiplash-

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Jan 22, 2014
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Hi, I'm sorry but I need help with calculus again. I've been doing pretty good ever since I started doing graphing, but this question is has got me stumped as what to do next:

find the intervals of increase. use the first derivative test to find the local maxima and minima. sketch a rough graph.

F(x) = (x2-1)/(x2+1)

I was able to find the derivative (I think) using the quotient rule:

F'(x)=(2x)/(x4+1)

and I set F'(x) to zero to find the critical point, but I don't know how to find X from a fraction like this. what should I do?

0=(2x)/(x4+1)

x=?
 
Hi, I'm sorry but I need help with calculus again. I've been doing pretty good ever since I started doing graphing, but this question is has got me stumped as what to do next:

find the intervals of increase. use the first derivative test to find the local maxima and minima. sketch a rough graph.

F(x) = (x2-1)/(x2+1)

I was able to find the derivative (I think) using the quotient rule:

F'(x)=(2x)/(x4+1)

and I set F'(x) to zero to find the critical point, but I don't know how to find X from a fraction like this. what should I do?

0=(2x)/(x4+1)

x=?
Let's start with the quotient rule, which you need to memorize.

\(\displaystyle u(x)\ and\ v(x)\ are\ differentiable\ functions\ and\ f(x) = \dfrac{u(x)}{v(x)} \implies f'(x) = \dfrac{u'(x)v(x) - u(x)v'(x)}{\left\{v(x)\right\}^2}.\)

I have found that doing things in small pieces tends to reduce errors.

In this case, \(\displaystyle u(x) = x^2 - 1 \implies u'(x) = 2x\ so\ u'(x)v(x) = what?\)

And in this case \(\displaystyle v(x) = x^2 + 1\ so\ v'(x) = what,\ u(x)v'(x) = what,\ and\ \left\{v(x)\right\}^2 = what?\)

Now put all that together to find the derivative of f(x). What is it?

If f'(x) = 0, the numerator of the derivative must equal what? Can you solve that equation.

So where is f'(x) = 0? Where is f'(x) negative? And where is f'(x) positive?

Can you solve the problem now?
 
Thanks, that actually made me notice an error, since I screwed up the power rule (somehow it slipped my head but anyway):


u(x) = x2-1
u'(x) =2x
v(x) = x2+1
v'(x) = 2x

therefore:

f′(x)=[(2x)(x2+1)−(2x)(x2-1)]/[(x2+1)]2

f′(x)= (4x)/(x4+1)

My problem wasen't that is it was:

0=(4x)/(x4+1)

How does one find what X is from that? It's probably a really simple answer but I don't know how. with the other question I could factor or re arrange the equation but how would I find x for this? (is this actually an algebra question?)
 
Thanks, that actually made me notice an error, since I screwed up the power rule (somehow it slipped my head but anyway):


u(x) = x2-1
u'(x) =2x
v(x) = x2+1
v'(x) = 2x

therefore:

f′(x)=[(2x)(x2+1)−(2x)(x2-1)]/[(x2+1)]2

f′(x)= (4x)/(x4+1)

My problem wasen't that is it was:

0=(4x)/(x4+1)

How does one find what X is from that? It's probably a really simple answer but I don't know how. with the other question I could factor or re arrange the equation but how would I find x for this? (is this actually an algebra question?)
You are still not good on the derivative.

\(\displaystyle (x^2 + 1)^2 \ne x^4 + 1.\) It's easy in calculus to forget your algebra, but it will KILL you on a test.

I gave you a hint on the remaining part of the problem as well. \(\displaystyle \dfrac{a}{b} = 0 \implies a = 0.\)

With that extra hint, take a look again at my previous post. You are almost there.
 
You are still not good on the derivative.

\(\displaystyle (x^2 + 1)^2 \ne x^4 + 1.\) It's easy in calculus to forget your algebra, but it will KILL you on a test.

I gave you a hint on the remaining part of the problem as well. \(\displaystyle \dfrac{a}{b} = 0 \implies a = 0.\)

With that extra hint, take a look again at my previous post. You are almost there.


OHHHHHHHHHHHH

(x2+1)2 is the same as (x2+1)(x2+1)

Which would be:

(x4+2x2+1)

I completely forgot. I do tend forget basic math when doing this kind of stuff early on. Thank you.

I never thought that as long as the top part is 0, it doesn't matter what the bottom part of the fraction is. Thanks. okay, for the second part.
 
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