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How do you sketch the graph 1/4 X^4 - 4/3 X^3 -3 Show each step please.

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vagyika | Salutatorian

Posted May 29, 2013 at 9:53 PM via web

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How do you sketch the graph 1/4 X^4 - 4/3 X^3 -3

Show each step please.

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crmhaske | College Teacher | (Level 3) Associate Educator

Posted May 30, 2013 at 12:52 AM (Answer #1)

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First, solve for the roots of the function:

`f(x)=1/4x^4-4/3x^3-3=0`

`=1/12(3x^4-16x^3-36)`

`3x^4-16x^3-36=0`

There are a variety of methods that you can use to determine the roots of a polynomial not of degree 2, some more complex than others. 

Using Descarte's sign rule, we can see that there is a maximum of one positive root as the sign changes only once as we move from terms left to right (+ - -).

To determine how many negative roots there are compute f(-x):

`f(-x)=3x^4+16x^3-36`

Again, there is only one sign change (+ + -), so therefore there is a maximum of one negative root.

For simplicity, we will use an iterative method to solve for the roots of the function.  This simply means using trial and error.  We will solve for the negative root first.  Start with x=-1 and decrease by -1 until the sign of f(x) changes:

`f(-1)=3(-1)^4-16(-1)^3-36=-17`

`f(-2)=3(-2)^4-16(-2)^3-36=140`

Therefore the negative root is between -1 and -2:

f(-1.5)=33.19; f(-1.25)=2.57; f(-1.125)=-8.41

Oops, we've gone too far.  The negative root is between -1.25 and -1.125:

f(-1.2)=-2.13; f(-1.21)=-1.22; f(-1.22)=-0.3

You can keep going until you get even closer to 0, but we'll stop here at two decimal places.  The negative root of the function is approximately x=-1.22.

Now we do the same for the positive root:

`f(1)=3(1)^4-16(1)^3-36=-49`

f(2)=-116; f(3)=-225; f(4)=-292; f(5)=-161; f(6)=396

The positive root is between 5 and 6:

f(5.5)=47.19; f(5.4)=-4.5; f(5.41)=0.42

Therefore, the positive root of the function is approximately x=5.41.

Now we take the derivative of the function in order to find the maximum and minimum points:

`f'(x)=1/12(12x^3-48x^2)=x^3-4x^2=0`

`x^2(x-4)=0`

Therefore there are maximum or minimum points at x=0 and x=4.

We can determine whether they are maximums or minimums by taking the second derivative of the function:

`f''(x)=3x^2-8x`

f''(0)=0: inconclusive, graph is flat at x=0

f''(4)=16>0: concave up, therefore a minimum

We can also use the second derivative to determine where there are a inflection points:

`x(3x-8)=0`

x=0 and x=2.67

Since we know at x=4 the function is concave up, then at x=2.67 the function changes from concave down to concave up.  And since we know that at x=2.67 the function is concave up, then at x=0 it switches from concave up to concave down.  Now we have all of the information we need to graph the function.

Here is the final graph.  To sketch by hand you would plot all of the points we calculated: the roots, the maxima/minima, inflection points, and then you would draw the graph based on what you know about the concavity between points.

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