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, d­²y/dx² = 0.    Be sure to find the sign (+ or -) of dy/dx and of d²y/dx² at all X values. Reference Lesson 13 and the text Appendix A (pp 694 – 698), as needed. 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. Attach work to convince me you carried out these calculations. An Excel spreadsheet can make calculations easier. If used, please attach the spreadsheet file and upload it with the rest of your work so that I can examine your formulas. The beginning and ending X values below are not to be considered critical values.   Write the first derivative (dy/dx) and the second derivative (d²y/dx²₎ you used or you will not receive credit for them.

                                                Y = X³ +3X² -9X +10

                                                                                                                                                           

X	-4	-3	-2	-1	0	1	2
Y
dy/dx
d2y/dx2
Label Point (MAX, MIN, INF)

. Use the seven X values and their Y values you found above (which include the critical values) to help neatly sketch the graph of this polynomial function over the range of X values given.   Alternatively use a spreadsheet to plot it. Your sketch must be consistent with the tabled values above (which means, if you claim a certain X value is a maximum, then the graph of it should show this same value as a maximum. Similarly, if you claim an X value is an inflection point, then the graph of it should show it to be so. A minimum should graph as a minimum, too. The point is, if you figure out how the derivatives SIGNAL which X values are critical points, the graph of the values should show them as such.)

Answer 1

Y = X³ + 3X²- 9X + 10

dY/dX = 3X² + 6X - 9

d²Y/dX² = 6X + 6

1	B	C	D	E	F	G	H	I
2	X	-4	-3	-2	-1	0	1	2
3	Y	30	37	32	21	10	5	12
4	dY/dX	15	0	-9	-12	-9	0	15
5	d2Y/dX2	-18	-12	-6	0	6	12	18

=3*C2^2+6*C2-9 is the formula to calculate dY/dX for X=-4 in excel. Drag it rightward t get the remaining values.=C2^3+3*C2^2-9*C2+10 is the formula to calculate Y for X=-4 in excel. Drag it rightward t get the remaining values.

=6*C2+6 is the formula to calculate d²Y/dX² for X=-4 in excel. Drag it rightward t get the remaining values.

{NOTE: The values of Y, dY/dX, and d²Y/dX² are calculated by plugging in the values of X in the respective equations}

For X=-3 Y is MAXIMUM i.e. at Y=37. Here, at this point d²Y/dX² < 0. So, Second Order Condition is fulfilled.

For X=1 Y is MINIMUM i.e. at Y=5. Here, at this point d²Y/dX² > 0. So, Second Order Condition is fulfilled.

INFLECTION POINT is defined as the point where change in direction of curvature occurs. Thus, d²Y/dX² = 0

Here, at X = -1, d²Y/dX² = 0 . So, INFLECTION POINT IS AT X = -1.

EXCEL PLOT { note: Use scatter plot in excel to do the same}

MY SKETCH