Andrew S. answered • 05/19/21
B.S. in Electrical Engineering
Hi Jennifer,
This question is asking for the local maximum and local minimums, this means that the parts of the graphs that go on into infinity are not what we care about in this type of problem.
For polynomials with higher orders in general, they may have up to 1 less than the highest order number of maximums or minimums. This means that potentially functions with x3 as the highest order will have at most 2 maximums or minimums and x4 functions will have at most 3 maximums or minimums.
the easiest trick to find these is to use a graphing calculator from an online tool or a physical handheld one. This is not likely the case for educational purposes so here is how to use calculus to find these values. The key word for this is "derivative test" for googling if you want more formal definitions.
for the first one f(x) = x3 - 3x + 2
1.) take the first derivative
f ' (x) = 3x2 - 3
2.) set that new equation to 0 and solve for x
0 = 3x2 - 3
3 = 3x2
1 = x2
x = ±1
This is the location of where the functions first derivative equals 0. Likely good candidates for the maximum and minimum locations. Which one is the max and which is the min without seeing the graph?
3.) Take second derivative of the function
f " (x) = 6x
4.) plug in the values where the function has an max or minimum.
f " (-1) = 6 (-1)
f " (-1) = - 6
and
f " (1) = 6 (1)
f " (1) = 6
for this part we only care if the answer is positive or negative, for positive results the graph is shaped like a U around the x point plugged in. For negative results the graph is shaped like an Π (upside down U). So this gives information about the shape at the location, this does not tell us if the point is a maximum or minimum.
For that we plug the x points we found into the original equation. This yields:
f (1) = 0
f (-1) = 4
the coordinates of the max or min is (1,0) and (-1, 4) these are the locations of the max and min. Considering the shape of the U is at the 1 location from the positive 6 and the shape of the Π is at the -1 location from the -6 the maximum is located at (-1,4) and the minimum is located at (1,0).
This may not always be the case so be sure to plug the values back into the original equation and check that the y output max and min values make sense, so 4 > 0 and is shaped like a hill so it is the maximum. 0 < 4 and is shaped like a valley so it is the minimum.
quick logic for the second part verifying these steps:
1.)
f(x) = x4 - 8x2 + 3
f ' (x) = 4x3 - 16x
2.)
0 = 4x3 - 16x
0 = x ( 4x2 - 16)
0 = 4x ( x2 - 4)
0 = 4x (x + 2) (x - 2)
so x = 0 , x = 2 , x = - 2
three possible points maxes or mins
3.)
f ' (x) = 4x3 - 16x
f " (x) = 12x2 - 16
f " (0) = 12(0)2 - 16
f " (0) = 12(0)2 - 16
f " (0) = - 16 shaped Π
f " (2) = 12(2)2 - 16
f " (2) = 12*4 - 16
f " (2) = 32 shaped U
f " (-2) = 12x2 - 16
f " (-2) = 32 shaped U
4.) plug points back in
f (x) = x4 - 8x2 + 3
f (0) = 0 - 0 + 3
f (0) = 3 (0 , 3) shaped Π
f (x) = x4 - 8x2 + 3
f (2) = 16 - 32 + 3
f (0) = -13 (2 , -13) shaped U
f (x) = x4 - 8x2 + 3
f (- 2) = 16 - 32 + 3
f (0) = -13 (- 2 , -13) shaped U
here 3 > - 13 and that means that (0 , 3) is a maximum of the function. Since the other two points share a minimum value, they are both the minimum for the function.
so one max at (0, 3) and two min at (2,-13) and (-2,-13)
side note, if you care about the inflection points set the second derivative to equal 0 and solve, this will tell you where the function changes from a min to a max or max to a min.
I hope this helps!
