# How do you find all values of x such that f(x) = 0 given f(x) = (1/4)x^3 - 2?

Jun 1, 2018

Assuming $f : \mathbb{R} \to \mathbb{R}$, then $x = 2$ is the only value which satisfies $f \left(x\right) = 0$.

#### Explanation:

We wish to find the roots of the function

$f \left(x\right) = \left(\frac{1}{4}\right) {x}^{3} - 2$

By root, we mean any value of $x$ for which $f \left(x\right) = 0$.

$\therefore \left(\frac{1}{4}\right) {x}^{3} - 2 = 0 \implies \left(\frac{1}{4}\right) {x}^{3} = 2$

Multiply both sides by $4$.

${x}^{3} = 8 \implies \textcolor{red}{x = 2}$

Let this first root we found be ${x}_{1}$. However, we did it algebraically and there seems to be only one solution. Could this be the only one?

Yes. If the function has domain and range over the real numbers $\mathbb{R}$ then $x = 2$ is the only root of $f$.

We can verify this graphically, too:

graph{(y-1/4x^3+2)=0 [-7.9, 7.9, -3.31, 4.59]}