# Graphing with calculus

I am attempting to graph $y = x^4 - 3x^3 + 3x^2 - x$

The derivative is $y=4x^3 -9x^2 + 6x-1$.

The second derivative is $y=12x^2-18x+6$.

The intercepts will be $x= 0$ and $y= 0$.

Past this, I can't find any way to get critical numbers. I do not know how to factor something like that all but I know it has roots.

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For the first derivative, you need to factor the expression $4x^3-9x^2+6x-1$. Towards this end, first try guessing a root of $4x^3-9x^2+6x-1$. The so called "rational roots theorem' will be useful here. Eventually, you might try $x=1$; and this is indeed a root of $4x^3-9x^2+6x-1$ as substitution will verify.

Now $x=1$ is a root of $4x^3-9x^2+6x-1$ if and only if $x-1$ is a factor of $4x^3-9x^2+6x-1$. This tells you that $4x^3-9x^2+6x-1$ factors as $$4x^3-9x^2+6x-1 = (x-1)(ax^2+bx+c)$$ for some constants $a$, $b$, and $c$. You can figure those out by doing the long division $${4x^3-9x^2+6x-1\over x-1}= 4x^2-5x+1.$$ So $$4x^3-9x^2+6x-1=(x-1)(4x^2-5x+1).$$

And now you need to factor $4x^2-5x+1$: $$4x^2-5x+1=(4x-1)(x-1)$$

So $f'(x)=(x-1)(x-1)(4x-1)$.

For the second derivative, you could use the quadratic formula to find its zeroes, or factor it: $$12x^2-18x+6=6(2x^2-3x+1)=6(2x-1)(x-1)$$

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I do not follow how you know that x-1 will be a factor. I do not know how to do long division either. How did you know how to factor $4x^2-5x+1$? Did you just sort of guess, trial and error? I do not really have the quadratic formula memorized, I have a really hard time with memorizing long formulas like that but I will try. – user138246 Apr 9 '12 at 3:11
@Jordan I added some extra information to address your first question (more generally $x=a$ is a root of the polynomial $p$ if and only if $x-a$ is a factor of $p$). this may help for the long division. I factored $4x^2-5x+1$ more or less by guessing. You should at least have the quadratic formula written down somewhere :) – David Mitra Apr 9 '12 at 3:15
@Jordan: Also, $4x^2 - 5x + 1 = 4x^2 - 4x - x + 1 = 4x(x - 1) - (x - 1) = (4x - 1)(x-1)$ - so this can be done naively. – mixedmath Apr 9 '12 at 3:17
@Jordan Yes, that works of course. You should be able to do something similar for $2x^2-3x+1$ (or even $12x^2-18x+6$). – David Mitra Apr 9 '12 at 3:20
@Jordan The root is (or might be) $p/q$. In this case your guesses should of the form $p/q$, where $p$ is $1$ or $-1$ (a divisor of $1$) and $q$ is one of $1,-1,2,-2,4,-4$ (a divisor of 4). So the only rational numbers that might be roots are $\pm1,\pm{1\over2},\pm{1\over4}$. – David Mitra Apr 9 '12 at 3:39

One way to proceed would be to note that $x^4 - 3x^3 + 3x^2 - x = x(x-1)^3$

Alternatively, note that $1$ is a root of the derivative. You can divide your derivative now to get the other 2 critical numbers.

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How would I know to factor is out like that? – user138246 Apr 9 '12 at 3:12
@Jordan: Well, it's pretty clear that you can factor out an $x$. Then, you either know the expansion of $(x-1)^3$ or you don't. But it's one of those forms that come up all the time, just like $(a-b)(a+b) = a^2 - b^2$, so perhaps you'd recognize it. – mixedmath Apr 9 '12 at 3:15
I have the difference of squares memorized but I forgot about the difference of cubes? Or whatever it is. – user138246 Apr 9 '12 at 3:16

The second derivative factors nicely. Note that you can take a $6$ out of each term. So, you can write it as: $$6 (2x^{2} - 3x +1)$$ $$6 (x-1)(2x-1)$$

From that, I would note the inflection points by setting $x - 1 = 0$ and $2x - 1 =0$. From there, you can use the second derivative test and plug in a few test points and check for concavity (i.e. is the graph "up like a cup" or "down like a frown"?). From there, it should be easy - let me know if you have any issues.

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1. Plot several points (near a root if you have a guess where one is if not go x=-1, 0, 1)
2. Find the roots of first derivative to find the local mins and maxes (remember when its positive the function is increasing and when its negative its decreasing)
3. Find the roots of the second derivative to find the points of inflection (remember when its positive the function is concave up like a bowl, and when its negative its concave down)

Note: You can use completing the square or the quadratic equation for finding the roots of the second derivative. For the first derivative do David suggested.

This should be enough to graph the function.

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