How do you graph #f(x)=x^4-3x^2+2x#?

Question

Ezekiel Alexander · Accepted Answer

I assume that you want to graph this without technology.

The graph is at the end of the solution. (I believe it's more instructive to not have it to start.)

#f(x)=x^4-3x^2+2x #

#f# is a polynomial, so, of course its domain is the set of all real numbers.

#y#-intercept #f(0)=0# The #y#-intercept is #0#. (Or #(0, 0)# if you prefer.)

#x#- intercept(s) : Solve #f(x)=x^4-3x^2+2x=0#

Factor out the #x#: to get #x(x^3-3x+2)=0# By inspection or by the Rational Zero Theorem (or "Test"), #1# is a zero of #(x^3-3x+2)#. By the Factor Theorem, #(x-1)# is a factor. Use division (in some form) to get the quadratic factor: #(x^3-3x+2) = (x-1)(x^2+x-2)#

The quadratic is straightforward to factor: #(x^2+x-2)=(x+2)(x-1)#

So we see that #f(x)=x(x+2)(x-1)^2# the zeros (the #x#-intercepts are: #-2, 0, 1# (#1# is a multiple zero of even multiplicity.)

(Write the intercepts: #(-2,0), (0,0)#, and #(1, 0)# if you prefer.)

Analysis of #f'# Although we can work with #f(x)# in any form, I prefer the standard polynomial form over the 3 factor form.

#f(x)=x^4-3x^2+2x# So #f'(x)=4x^3-6x+2# Which is never non-existent, so we only need to solve: #f'(x)=4x^3-6x+2 = 0# By inspection or by the Rational Zero Theorem or by observing that multiple zeros of a polynomial are also zeros of the derivative, we see that #1# is again a zero, so #(x-1)# is a factor and:

#f'(x)=4x^3-6x+2 = 2(2x^3-3x+1)=2(x-1)(2x^2+2x-1)#

The quadratic factor has irrational zeros: #(-1 +- sqrt3)/2#.

Observe that #1< sqrt 3 <2#, so one of the zeros is negative and the other positive. And #(-1 + sqrt3)/2 < (-1 + 2)/2 = 1/2 <1#.

The critical numbers for #f# are Left to right : #(-1 - sqrt3)/2, (-1 + sqrt3)/2," and " 1#.

For ease of reference, let's call the negative critical number #z_1 =(-1 - sqrt3)/2# and the positive one #z_2 = (-1 + sqrt3)/2#

We need to investigate the sign of #f'# on each of the intervals:

#(-oo,z_1)#, #(z_1, z_2,)#, #(z_2,1)#, #(1,oo)#

If you like test numbers, I'd suggest: #-10, 0, 1/2, 10#. It may not be clear that #1/2# is in #(z_2,1)# until it is observed that:

#sqrt3 < 2# #=># #(-1 + sqrt3)/2 < (-1 + 2)/2 = 1/2#

If you prefer, use a factor table.

Whichever method you use, you'll find that:

Increasing/Decreasing #f'(x) < 0# on #(-oo,z_1)#, so #f# is decreasing on #(-oo,z_1)# #f'(x) > 0# on #(z_1, z_2,)#, so #f# is increasing on #(z_1, z_2,)# #f'(x) < 0# on #(z_2,1)#, so #f# is decreasing on #(z_2,1)# #f'(x) > 0# on #(1,oo)#, so #f# is increasing on #(1,oo)#

Local extrema: #f(z_1) = f((-1 - sqrt3)/2)# is a local minimum (also global) #f(z_2) = f((-1 + sqrt3)/2)# is a local maximum #f(1) = 0# is a local minimum.

Analysis of #f''#

#f''(x)=12x^2-6 = 6(2x^2-1)#

Whose zeros are : #+- 1/sqrt2 = +-sqrt 2/2#

Investigating the sign of #f''# on the appropriate intervals leads us to:

Concavity: #f''(x) > 0# on #(-oo, -sqrt2 / 2)# So #f# is concave up. #f''(x) < 0# on #(-sqrt2 / 2, sqrt2 / 2)# So #f# is concave down. #f''(x) > 0# on #(sqrt2 / 2, oo)# So #f# is concave up.

There are two inflection points. They are: #((-sqrt2 / 2, f(-sqrt2 / 2) )# which is #(-sqrt2 / 2 "," -5/4-sqrt2 )# and #((sqrt2 / 2, f(sqrt2 / 2) )# which is #(sqrt2 / 2 "," -5/4+sqrt2 )#

Now sketch the graph (It may take a couple of rough sketches first:

graph{y=x^4-3x^2+2x [-10, 10, -5, 5]}

Charles Anctil · Answer

undefined To graph the function $ f(x) = x^4 - 3x^2 + 2x $, you can follow these steps:

1. Find the critical points by setting the derivative of the function equal to zero and solving for $ x $.
2. Determine the intervals where the function is increasing or decreasing by analyzing the sign of the derivative.
3. Find the local maximum and minimum points, if they exist, by analyzing the sign changes in the derivative.
4. Determine the intervals where the function is concave up or concave down by analyzing the sign of the second derivative.
5. Find the inflection points, if they exist, by analyzing the sign changes in the second derivative.
6. Use this information to sketch the graph of the function.

Would you like to see the detailed calculations for these steps?