Expert: Scotto - 3/28/2011

Question
(a) Find the x- and y-intercepts of the function.
(b) Draw a number line for f:
(c) Find all critical numbers for f0:
(d) Draw a number line for f0:
(e) Find all zeros of f00:
(f) Draw a number line for f00:
(g) Find all in‡ection points of f:
(h) Sketch the graph of f. Label all in‡ection points. Label your axes.

Answer
Looking at the coefficients, they are 1 -3 1 1, and the sum of those numbers is 1.
That means that (x-1) is a factor.  Taking (x-1) out front gives (x-1)(x²-2x-1).

Using the quadratic equation, the other roots are (2+√(4+4))/2 and (2-√(4+4))/2.
These simplify to 1+√2 and 1-√2.

(a) This would mean the x-intercepts would be at x=1-√2, x=1, and x=1+√2.
The y-intercept is as f(0), and that can be found.

(b) To answer this, I would use -3, -2, -1, 0, 1, 2, 3 for the x values, which are put on the left side in seven rows and then find f(x) for each of these value for the right side.
For example, the first row would be -3 and f(-3).  It can be seen that
f(-3) = (-3)^3 - 3(-3)^2 - 3 + 1 = -27 - 27 - 3 - 1 = -58, so the first row is -3   -58.

(c) I'm not sure, but I think the critical number are the solutions and y-intercept,
which have been given.

(d) What is a, "number line for f0"?

(e) The zero values were given up at the top as 1-√2, 1, and 1+√2.

(f) What is, "a number line for f00"?

(g) Is, "in‡ection" for, "intersection"?
The intersection points are the x and y intercepts, which were given.

(h) Using the above values, plot where the graph crosses the x and y axis,
then draw a curve the fits.  It might be good to know that the max/min values of the function occurc at where 3x^2 - 6x + 1 = 0.  That is, x = (6-√(36-12))/6 and (6+√(36-12))/6.
These x values are 1 - (√6)/3 and 1 + (√6)/3.  The y value of the 1st one is a maximum at a little over y=1 and the y value for the 2nd one is a little under y=-1.