Q.Show that there is a root for the equation $x^3 - 3x = 0$ between 1 and 2.

The Intermediate Value Theorem states that if $f$ is continuous on a closed interval $[a,b]$, then for any value $y_0$ lying between $f(a)$ and $f(b)$, there exists at least one point $c$ in $[a,b]$ such that $f(c)=y_0$. Geometrically, this means that any horizontal line drawn at a height strictly between $f(a)$ and $f(b)$ must cross the curve $y=f(x)$ at least once somewhere over the interval — a continuous curve simply cannot skip over an in-between height on its way from one endpoint to the other. Continuity on the whole closed interval is essential to the theorem: if $f$ were discontinuous at even a single point inside $[a,b]$, the graph could genuinely jump straight over a value without ever reaching it, and the conclusion could fail. The theorem's most common use is to prove that an equation $f(x)=0$ has a root inside a given interval, without needing to solve the equation exactly: it suffices to show $f$ is continuous there (true automatically for any polynomial, or any other function built from the standard continuous families) and that $f$ takes a negative value at one endpoint and a positive value at the other, so that $y_0=0$ lies strictly betw …