Posted by Andrew on November 07, 2002 at 17:06:34:
In Reply to: Re: Find critical points? posted by T.Gracken on November 07, 2002 at 15:35:53:
: : I'm having a lot of trouble on my calculus homework right now. Here are the problems:
: : USE FUNCTION: f(x) = 3x^4 - 4x^3 + 6
: : 1) Find all critical points of y = f(x). Identify each one by its x coordinate.
: : 2) Use the first derivative test to determine each critical point of f(x) whether it is a local minimum, local maximum, or neither.
: : thanks
: BE CAREFUL HERE... the definitions I use are not always the same as others use.
: I define a critical point as a point of a function for which the graph of the function has either a horizontal tangent or an undefined tangent.
: The x-coordinates of critical points can be determined by
: 1. evaluating the derivative of the function, and
: 2. determining the values for which the derivative is either equal to 0 or undefined.
: So, for f(x)= 3x4 - 4x3 + 6
: 1. f '(x) = 12x3 - 12x2
: 2. 12x3 - 12x2 = 0 when x=0 or x=1. [there are no values of x for which this expression is undefined]
: the critical points of the function then have x-coordinates of 0 and 1.
: As to the first derivative test...
: draw a numberline and plot the critical numbers found in step 2 (above).
: choose a value in the intervals created on the numberline and (a) if the number causes the derivative to be positive, then draw an upwards arrow over that interval [indicating the graph rises (increases) on that interval], (b) if the number causes the derivative to be negative, then draw a downwards arrow over that interval [indicating the graph falls (decreases) on that interval]. Then read what the first derivative is from your text and interpret the picture.
: That is, if the graph increases on the left of the critical number and decreases on the right, the critical number is at a maximum.
: if the graph decreases on the left and increases on the right then there will be a minimum.
: if the graph does not change (i.e. both sides are the same) then neither a max or min is there.
: With the numberline, it is very easy to interpret the results with just a few practice problems.
: -------- 0 ------- 1 --------- (number line)
: pick a number less than 0 (i will use -1). sub into derivative and you get -24 [negative number so draw down arrow over left portion of numberline]
: pick a number between 0 and 1 (i will use 1/2). sub into derivative and get -3/2 [negative number so draw down arrow over middle portion of numberline]
: pick a number bigger than 1 (i will use 2). sub into derivative and get 48 [positive number so draw up arrow over right portion of numberline]
: notice the only place arrows change direction are from middle portion to right portion.
: interpretation of first derivative test:
: when x=0 there is neither max nor min,
: when x=1 there is a minimum.
There are 2 more problems with the homework that go along with what you did above.
3) When the domain is restricted to -2 <= x <= 2, what is the global maximum and global minimum value of f(x) over this domain? Name both the value of the function and the x coordinate where the extreme value occurs.
4) Using Xmin=-2 and Xmax=2 and appropriate values of Ymin and Ymax, sketch a useful graph of y=f(x). Be sure to label with both the x and y coordinates all local extrema, global extrema, and inflection points.
THANK YOU! THIS IS DUE TOMORROW AND I DO NOT KNOW WHAT TO DO ON THESE LAST 2!!!
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