Question

show that f(x) = (x^3)+(3x)-(1)=0, has a root between (0,1)​

Answer 1

Here,

$\longrightarrow\sf{f(x)=x^3+3x-1}$

Its derivative with respect to $\sf{x}$ is,

$\longrightarrow\sf{f'(x)=3x^2+3}$

We see this derivative is always positive for every $\sf{x\in\mathbb{R}.}$ Hence $\sf{f(x)}$ is a strictly increasing function.

$\longrightarrow\sf{x^2\in[0,\ \infty)}$

$\longrightarrow\sf{3x^2\in[0,\ \infty)}$

$\longrightarrow\sf{3x^2+3\in[3,\ \infty)}$

$\Longrightarrow\sf{f(x)\ \textgreater\ 0}$

For $\sf{x=0,}$

$\longrightarrow\sf{f(0)=(0)^3+3(0)-1}$

$\longrightarrow\sf{f(0)=-1}$

For $\sf{x=1,}$

$\longrightarrow\sf{f(1)=(1)^3+3(1)-1}$

$\longrightarrow\sf{f(1)=3}$

That is, since $\sf{f(x)}$ is strictly increasing,

$\longrightarrow\sf{x\in[0,\ 1]\quad\implies\quad f(x)\in[-1,\ 3]}$

This implies $\sf{f(x)}$ intersects x axis at a point $\sf{(x,\ 0)}$ where $\sf{x\in(0,\ 1)}$ because $\sf{0\in[-1,\ 3].}$

Therefore, $\sf{f(x)=x^3+3x-1=0}$ has a root in the interval $\sf{(0,\ 1).}$