HELP (inflection points and whatnot)

• bilderback
In summary, the conversation includes discussions on differentiating equations, finding absolute maximum and minimum values, and determining concavity and points of inflection for trigonometric functions on specific intervals. The process for solving these problems involves taking derivatives and setting them equal to 0 to find critical points, and using the sign of the second derivative to determine concavity.
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)

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.

What are inflection points?

Inflection points are points on a graph where the concavity changes from positive to negative or vice versa. They can also be thought of as points where the rate of change (slope) transitions from increasing to decreasing or vice versa.

How do you find inflection points?

To find inflection points, you must first take the second derivative of the function and set it equal to 0. Then, solve for the x-values that make the second derivative equal to 0. These x-values correspond to the inflection points on the graph.

What is the significance of inflection points?

Inflection points can indicate changes in the behavior of a function. They can help identify where a function transitions from increasing to decreasing or vice versa, and can also help identify where the curve changes from concave up to concave down or vice versa.

Are inflection points always present on a graph?

No, not all functions have inflection points. Functions that are linear (have a constant slope) or have a constant concavity will not have any inflection points. Additionally, some functions may have multiple inflection points.

Can inflection points occur at discontinuities?

No, inflection points cannot occur at discontinuities. Inflection points require the function to be continuous and differentiable at that specific point. Discontinuities indicate a break in the function, which prevents the existence of an inflection point.

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