For this post i am going to go into detail on the second derivative test to find all possible points of inflection and intervals of concavity. remember, points of inflection only happen where there is a change of concavity.
Example: f'(x)= 6/(x^(2)+3)
First, you have to take the derivative of that, and you have to use the quotient rule, so the beginning of the problem will look like, [(x^(2)+3)(0)-6(2x)]/(x^(2)+3)^(2), which simplifies to -12x/(x^(2)+3)^2 remember, that was just the first derivative.
Second step is to take the derivative of the first derivative, that would make this step called taking the second derivative.
Once again u need to use the quotient rule, so f''(x)={(x2+3)^2-(12)-[(-12x)2(x^(2)+3)2x} all that over (x^(2)+3)^4 then you get a bunch of stuff, then you simplify, then you cancel, so I am just going to type the end answer of the second derivative. Which is, (3)(6)(x+1)(x-1) all over (x^2+3)^3
The possible points of inflection are found in the numerator of the finished second derivative, in this case, if you look, it would be x=1, and x=-1
so then you set up your points, (-infinity, -1) u (-1, 1) u (1, infinity)
then you plug in. f''(-2)= positive value f''(0)=negative value f''(2)=positive value
then you know that your intervals concave up at (-infinity, -1) u (1,infinity) or x<1,>1
and it is concave down at (-1,1) or -1
and you're points of inflection are x=-1, and x=1
that's it for the second derivative test, the only thing i have any problems with is substition... so yeah..
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substitution is pretty easy...like you need to get better at recognizing when you do use substitution.
ReplyDeletebasically if there is something in the problem within parenthesis or set aside somehow...and in that problem somewhere else, there is the derivative of the above said. you use substitution
for example,
the integral of sin(3x+2)
your u would be 3x+2 and du would be 3. well 3 isn't in the problem so you would rewrite this as
1/3 the integral of sin(u)du
then you integrate as normal and then plug back in for u. simple as that.
You use substitution where there is a product rule or a quotient rule in an integral because you cannot do product or quotient rules while integrating. When looking at the problem, there will be two things being multiplied together. Find the one that's the derivative of the other. The one that's the "origional function" will be u and the derivative of that will be du. Integrate udu, then solve.
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