June Bryan

2023-03-12

How to find all zeros of $f\left(x\right)=3{x}^{3}-12{x}^{2}+3x$?

Karbamidjts

We must first determine the derivative of a function before we can determine its zeros. This one is simple; we'll just use the power rule to evaluate each phrase individually.
$f\left(x\right)=3{x}^{3}-12{x}^{2}+3x$
$f\prime \left(x\right)=9{x}^{2}-24x+3$
Then we set the derivative equal to zero and solve for x
$9{x}^{2}-24x+3=0$
Given that this function cannot be factored, we can enter this into our calculator to determine the two x values. Obtaining two values
$x=2.535$ $x=.131$

udahnulizwk

First, factor out an x:
$f\left(x\right)=x\left(3{x}^{2}-12x+3x\right)$
Already, you can figure out that when x zero, f(x) is zero, so $x=0$ is one solution.
Then, attempt to factor the quadratic. Unfortunately, in this case, you can't use factoring, so plug the numbers into the quadratic formula, which is $x=\frac{-b±\sqrt{{b}^{2}-4ac}}{2}a$ in case you don't have it memorized:
$x=\frac{-\left(-12\right)±\sqrt{{\left(-12\right)}^{2}-4\left(3\right)\left(3\right)}}{2\left(3\right)}$
$x=\frac{12±\sqrt{144-36}}{2\left(3\right)}$
$x=\frac{12±\sqrt{108}}{6}$
(The square root can be simplified to $6\sqrt{3}$, because $\sqrt{108}=\sqrt{6\cdot 6\cdot 3}={\sqrt{6}}^{2}\cdot \sqrt{3}=6\cdot \sqrt{3}$)
$x=\frac{12±6\sqrt{3}}{6}$
A six can also be cancelled out:
$x=\left(2±\sqrt{3}\right)$
That means the zeroes are $x=0,x=2+\sqrt{3},\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}x=2-\sqrt{3}$

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