Question

Calculate the Y values corresponding to the X values given below. Find the critical values for X for the given polynomial by finding the X values among those given where the first derivative, dy/dx = 0 and/or X values where the second derivative, d¬2y/dx2 = 0. Be sure to indicate the sign (+ or -) of dy/dx and of d2y/dx2 tabled values. Using the first and second derivative tests with the information you have calculated, determine which X value(s) represent maximums (MAX), which minimums (MIN) and which inflection points (INF). Label the qualifying X value as such. The beginning (-.333) and ending X values (1) below are not to be considered critical values.

write the first derivative (dy/dx or Y’). Set this equal to zero and solve for the X values that make it equal to zero. Also write the second derivative (d2y/dx2 or Y”). Set this equal to zero and solve for the X values that make it equal to zero. Complete the table.

Y = X³ –X² +3

X	-.333	-.25	0	.25	.333	.667	1
Y
dy/dx
d2y/dx2
Label Point (MAX, MIN, INF)

Use the nine X values and their Y values you found above (which include the critical values) to help neatly draw the graph of this polynomial function over the range of X values given.  

Answer 1

For the function be , we have or and or or .

The table would be as below.

X	-0.333	-0.25	0	0.25	0.333	0.667	1
Y	+2.8522	+2.9219	+3	+2.9531	+2.926	+2.8519	+3
	+0.9987	+0.6875	0	-0.3125	-0.3333	0	+1
	-3.998	-3.5	-2	-0.5	0	2.002	4
Label Point	-	-	MAX	-	INF	MIN	-

The critical values would be where or or or or . For X=0, we have, ie , meaning that we have a maximum at X=0. For X=0.667, we have , ie , meaning that we have a minimum at X=0.667.

The inflection point would be where or or .

The graph would be as below.

As can be seen, the curve is maximum at X=0, at a minimum at X=0.667 and at an inflection point at X=0.333 (inflection point is marked as the red-dot, which is where Y's curvature is changed).