# Re: Find critical points?

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Posted by T.Gracken on November 08, 2002 at 13:11:13:

In Reply to: Re: Find critical points? posted by Andrew on November 07, 2002 at 17:06:34:

: : : 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.

: : so...

: : -------- 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.
start paying attention in class!!!

for number three the same thing is done, but instead of allowing the numberline to extend indefinitely (right and left), start it at -2 and end it at 2. Then use the fact that endpoints might also produce extreme values.
for number four, look at the graph that is produced, but start do not look left of when x is -2 and do not look right of x = 2. consider how high and low the graph goes within this section to determine y values.

: 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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