HELP (inflection points and whatnot)

[bilderback]

HELP! (inflection points and whatnot)

Okay I know how to differentiate but I suck at simplifying the equation

For example

f''(x) = (x^2 + 3)^2(-12)-(-12x)(2)(x^2+3)(2x)
^Second derivative

all over

(x^2+4)^4

Now the answer is 36(x^2-1) over (x^2+3)^3

But as to how they came up with 36(x^2-1) is beyond me, I tried breaking up the numerator and solving one half of the equation and the other half; breaking it down piece by piece, but I came up with an equation that resembled nothing like I was suppose to get.

The next question I have is finding the absolute max and min of f(x)= sq root of 81-x^2 in the interval (-9,9)

it says that it's (NONE, NONE) for an abs min and (0,9) for an abs max

then they want me to get it on an interval of [4.5,9) which leads to (NONE,NONE) for the abs min and (4.5, 7.89) for a max. How does that work?

As for the trig problems...I don't even know how to go about those...f(x)=8x+sin(x)

I know the derivative of this is f'(x)=8+cos(x) but finding the open intervals where the function is increasing/decreasing as well as applying the first derivative test to find the relative extrema is something I don't know how to go about doing...maybe I'm making it a little complicated...

I don't know how to find the concavity/points of inflection of f(x)=4csc(5x/2) on the interval (0,2pi) or how to find it with this function f(x)=8sin(x)+4cos(x)

 

[kurt.physics]

on the interval (0,2pi)For the first one, you can take the second derivative of f(x)=4csc(5x/2) and set it equal to 0. Then solve for x. This will give you the x-coordinates of any inflection points. For the second one, take the second derivative of f(x)=8sin(x)+4cos(x) and set it equal to 0. Then solve for x. This will give you the x-coordinates of any inflection points. You can then use the sign of the second derivative to determine the concavity of the function at those points.