# Timeless Theorems of Mathematics/Intermediate Value Theorem

The **Intermediate Value Theorem** is a fundamental theorem in calculus.
The theorem states that if a function, $f(x)$ is continuous on a closed
interval $[a,b],$ then for any value, $y$ defined between $(f(a)$ and
$f(b),$ there exists at least one value $c \in (a,b)$ such that
$f(c) = y$. ![Intermediate Value Theorem: f(x) is continuous on \[a,
b\], there exists at least one value c, that is defined on (a, b) such
that f(c) =
y.](Intermediatevaluetheorem.svg "Intermediate Value Theorem: f(x) is continuous on [a, b], there exists at least one value c, that is defined on (a, b) such that f(c) = y.")

## Proof

**Statement:** If a function, $f$ is continuous on $[a, b],$ then for
every $y$ between $f(a)$ and $f(b),$ there exists at least one value
$c \in (a, b)$ such that $f(c) = y$

**Proof:** Assume that $f(x)$ is a continuous function on $[a, b]$ and
$f(a)<f(b).$

Consider a function $g(x) = f(x) - y.$ The purpose of defining $g(x)$ is
to investigate the behavior of $f(x)$ concerning the value $y$.

Since $f$ is continuous on $[a,b]$ and $y$ is a constant,
$g(x) = f(x) - y$ is also continuous on $[a,b],$ as the difference of
two continuous functions is continuous.

Now, $f(a) < y$ \[As $c \in (a, b)$ and $y = f(c)$\]

Or, $f(a) - y < 0$

$\therefore g(a) < 0$

In the same way, $g(b) > 0$

Since $g(x)$ is continuous and $g(a)$ is defined below the $x$-axis
while $g(b)$ is defined above the $x$-axis, there must exist at least
one point $c$ in the interval $[a,b]$ where $g(c)=0$.

Therefore, at the point $c$, $g(c) = f(c) - y = 0 \implies f(c) = y$

∴ There exists at least one point $c$ in the interval $[a,b]$ such that
$f(c)=y.$ ***\[Proved\]***

