6) Take the derivative:
f'(x) = 3x^2 - 2x -1
Are there any real numbers that make f(x) undefined? If yes then the domain isn't all real, if no then the domain is all real.
f(x) is increasing where f'(x) is positive
f(x) is decreasing where f'(x) is negative
Solve for the values of x where the derivative = 0, these are the points where the function could possibly be transitioning between increasing/decreasing. Test values around these points to find if the derivative is positive or negative
Relative extrema/Local max and mins are going to be at locations where the derivative = 0, find these x-values and plug them back into f(x) to find the values
7) Take the first and second derivative
f'(x) = 3x^2 - 2x - 1
f''(x) = 6x - 2
The critical values are the x-values where f'(x) = 0
To find the intervals of concavity find the values where f''(x) = 0 and then plug in x-values around these points. If f''(x) is negative at a certain x-value then f(x) is concave down and if it is positive f(x) is concave up.
8) The absolute extrema of a function are the absolute max and min as opposed to the local max and min.
Take the derivative of the function.
f'(x) = 3x^2 - 2x - 1
Solve for the values of x where f'(x) = 0.
These values are possibly the locations of absolute extrema, take these values and plug them into f(x). To find the absolute extrema over and interval you also have to plug in the end points into f(x) to see if they are larger or smaller than the values where f'(x) = 0. You should now have 4 numbers, the largest and smallest are your two absolute extrema.
9) Simplify the function and you get ( x * (x + 2) ) / (x + 1)
The domain is all real numbers except x = -1 because then the function is undefined.
There is a hole at x = 2, we are able to divide this factor out when we simplify it, so it isn't an asymptote, but it still makes the original function undefined and it's thus a hole.
Plug x =0 into the function and you'll see the y-intercept is 0
Find the values where the function = 0 and you'll have your x-intercepts
The VA (Vertical Asymptote I assume?) occurs at x = -1 (Where it is undefined)
Not sure what an O Asymptote is?
Take the derivative.
Solve for where it is negative or positive to find increasing/decreasing
Find the values where the derivative = 0 to find locations of relative extrema
Find where the 2nd derivative is positive or negative to find it's concavity
Find where the 2nd derivative = 0 to find points of inflection
This post was edited by RzChaos on May 26 2015 06:42pm