Re: Find critical points?

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: : : 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. : 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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