June Bryan

2023-03-12

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

Karbamidjts

Beginner2023-03-13Added 4 answers

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

$x=2.535$ $x=.131$

$f\left(x\right)=3{x}^{3}-12{x}^{2}+3x$

$f\prime \left(x\right)=9{x}^{2}-24x+3$

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

Beginner2023-03-14Added 5 answers

First, factor out an x:

$f\left(x\right)=x(3{x}^{2}-12x+3x)$

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 $x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2}a$ in case you don't have it memorized:

$x=\frac{-(-12)\pm \sqrt{{(-12)}^{2}-4\left(3\right)\left(3\right)}}{2\left(3\right)}$

$x=\frac{12\pm \sqrt{144-36}}{2\left(3\right)}$

$x=\frac{12\pm \sqrt{108}}{6}$

(The square root can be simplified to $6\sqrt{3}$, because $\sqrt{108}=\sqrt{6\cdot 6\cdot 3}={\sqrt{6}}^{2}\cdot \sqrt{3}=6\cdot \sqrt{3}$)

$x=\frac{12\pm 6\sqrt{3}}{6}$

A six can also be cancelled out:

$x=(2\pm \sqrt{3})$

That means the zeroes are $x=0,x=2+\sqrt{3},\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}x=2-\sqrt{3}$

$f\left(x\right)=x(3{x}^{2}-12x+3x)$

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 $x=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2}a$ in case you don't have it memorized:

$x=\frac{-(-12)\pm \sqrt{{(-12)}^{2}-4\left(3\right)\left(3\right)}}{2\left(3\right)}$

$x=\frac{12\pm \sqrt{144-36}}{2\left(3\right)}$

$x=\frac{12\pm \sqrt{108}}{6}$

(The square root can be simplified to $6\sqrt{3}$, because $\sqrt{108}=\sqrt{6\cdot 6\cdot 3}={\sqrt{6}}^{2}\cdot \sqrt{3}=6\cdot \sqrt{3}$)

$x=\frac{12\pm 6\sqrt{3}}{6}$

A six can also be cancelled out:

$x=(2\pm \sqrt{3})$

That means the zeroes are $x=0,x=2+\sqrt{3},\phantom{\rule{1ex}{0ex}}\text{and}\phantom{\rule{1ex}{0ex}}x=2-\sqrt{3}$