Theorem: Suppose $f:(a, b) \rightarrow \mathbb{R}$ is a differential function.

(i) If $f^{\prime}(x)>0 \quad \forall x \in(a, b)$, then $f(x)$ is strictly increasing on $(a, b)$

(ii) If $f^{\prime}(x)<0 \quad \forall x \in(a, b)$, then $f(x)$ is strictly decreasing on $(a, b)$.

Proof: (i) Let $a<x_1<x_2<b$.

Then $f$ is cont. on

$\left[x_1, x_2\right]$ is diff. an

$\left(x_1, x_2\right)$.

$\therefore$ By the MVT, $\exists c \in\left(x_1, x_2\right)$ such that

$\begin{aligned}& \frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}=f^{\prime}(c)>0 \\ \Rightarrow & f\left(x_2\right)-f\left(x_1\right)>0 \quad\left(\because x_2-x_1>0\right) . \\ \Rightarrow & f\left(x_2\right)>f\left(x_1\right) \text { whenever } x_2>x_1 . \end{aligned}$

(ii) Similar to (i).

Problem-1:

Prove that $\sin x \leqslant x \quad \forall x \geqslant 0$.

Solution:$~$ Let $f(x)=x-\sin x$

$\text { Then, } ~ f^{\prime}(x)=1-\cos x \geqslant 0 \quad(\because ~ -1 \leq \cos x \leqslant 1)$

$\therefore f(x)$ is an increasing function. Also, $\quad f(0)=0$.

$\therefore$ For $x \geqslant 0, f(x) \geqslant f(0)=0$

ie. $x-\sin x \geqslant 0$

ie. $\sin x \leq x \quad \forall x \geqslant 0$.

Problem-2:

Show that $x<\tan x \quad \forall x \in\left(0, \frac{\pi}{2}\right)$

Solution: $~$ Put $f(x)=\tan x-x$

Then $f^{\prime}(x)=\sec ^2 x-1=\tan ^2 x>0$

$\therefore f(x)$ is strictly increasing $m\left(0, \frac{\pi}{2}\right)$

$\Rightarrow \quad f(x)>f(0)=0 \quad \forall x \in\left(0, \frac{\pi}{2}\right)$

ie. $\tan x>x \quad \forall x \in\left(0, \frac{\pi}{2}\right)$.

Prove that $|\sin x-\sin y| \leqslant|x-y|$ for all $x, y \in \mathbb{R}$.

Solve: $\quad f(x)=\sin x$ is cont. and differentiable everywhere.

$\therefore$ By the MVT, given $x<y, \exists c \in(x, y)$ such that $f^{\prime}(c)=\frac{f(y)-f(x)}{y-x}$

ie. $\frac{\sin y-\sin x}{y-x}=f^{\prime}(c)=\cos c$

$\Rightarrow\left|\frac{\sin x-\sin y}{x-y}\right|=|\cos x| \leq 1 \text { if } x \neq y$

$\Rightarrow \quad|\sin x-\sin y| \leqslant|x-y| .$

Note that continuity is required for the IVT.

Corollary: Suppose $f:[a, b] \rightarrow \mathbb{R}$ be continuous and assume that $f(a)$ and $f(b)$ are of opposite signs (ie. $f(a) f(b)<0$ ). Then $\exists$ at least one $x \in(a, b)$ s.t. $f(x)=0$.

Pf: since o lies between $f(a)$ & $f(b)$, by the IVT $\exists x \in(a, b)$ s.t. $f(x)=0$.

Applications of IVT:

Theorem: Every polynomial of odd degree must have at least one zero.

(ie. if $p(x)=a_0+a_1 x+\cdots+a_n x^n$ where $n$ is odd & $a_n \neq 0$ then $\exists$ at least one $c \in \mathbb{R}$ s.t. $\quad p(c)=0)$

Remark: The result is not true for even degree polynomials, e.g. $p(x)=x^2+1$ has no real zeros.

Pf of the them: $p(x)=a_0+a_1 x+\cdots+a_n x^n$

Since $n$ is an odd integer, $\lim _{x \rightarrow+\infty} x^n=+\infty$ and

$\lim _{x \rightarrow-\infty} x^n=-\infty \text {. }$

$\therefore p(x)$ must be negative at some $x$ and positive at some other $x$.

$\therefore$ &y th cor. to IVT, $p(x)$ must be zero at same point $x$.

Show that $x^5+4 x-1=0$ has exactly one solution in $\mathbb{R}$.

Solve:$~$ Let $f(x)=x^5+4 x-1$

Since $p(x)$ is an odd degree polynomial, $p(x)=0$ has at least one solution in $\mathbb{R}$.

Suppose there are two solutions $x_1<x_2$.

Then $p\left(x_1\right)=0=p\left(x_2\right)$.

By the Rolle's theorem, $\exists c \in\left(x_1, x_2\right)$ s.t. $p^{\prime}(c)=0 \quad \therefore p(x)=0$ has

But, $p^{\prime}(x)=5 x^4+4 \geqslant 4 \quad \therefore p(x)=0$ has

$\therefore$ We get a contradiction.

What is the solution. to $p(x)=0$ ?

Solution :

We can use the IVT to find approximate solution as follows.

$\begin{aligned}& p(x)=x^5+4 x-1 \\& p(0)=-1<0, p(1)=4>0\end{aligned}$

By the IVT, we know $p(c)=0$ for same $c \in(0,1)$ Find the value at the mid point $\frac{1}{2}$.

$p\left(\frac{1}{2}\right)=\frac{1}{32}+2-1=1+\frac{1}{32}>0$

Since $p(0)<0, p\left(\frac{1}{2}\right)>0$, the zero must lie in $\left(0, \frac{1}{2}\right)$.

$p\left(\frac{1}{4}\right)=\frac{1}{4^5}+1-1=\frac{1}{4^5}>0 \text {, so, ........... }\left(0, \frac{1}{4}\right)$

Now, $p\left(\frac{1}{8}\right)=\frac{1}{8^5}+\frac{1}{2}-1=\frac{1}{8^5}-\frac{1}{2}<0$

$\therefore$ The zero must lie in $\left(\frac{1}{8}, \frac{1}{4}\right)$

Proceeding this way we can get the zero lying in smaller & smaller interval.

This method is called the bisection method.

Prove that $\cos x=x$ for some $x \in\left(0, \frac{\pi}{2}\right)$.

Solution:$~$ Let $f(x)=\cos x-x$

$f(x)$ is continuous everywhere

Also, $f(0)=\cos 0-0=1>0$

$f\left(\frac{\pi}{2}\right)=\cos \frac{\pi}{2}-\frac{\pi}{2}=0-\frac{\pi}{2}=-\frac{\pi}{2}<0$

By the IVT, $\exists x \in\left(0, \frac{\pi}{2}\right)$ s.t. $f(x)=0$

$\text { ie. } \cos x=x \text {. }$

Problem: Let $f:[0,2] \rightarrow \mathbb{R}$ be continuous. Also, $f(0)=f(2)$

Prove that there exist $x, y \in[0,2]$ s.t. $y-x=1$ and $f(x)=f(y)$

Solution: $T_0$ find $x$ & $y$ st. $y=x+1$ and $f(x)=f(y)$

ie. we reed show the existence of $x \in[0,1]$ s.t. $f(x)=f(x+1)$

Let $g(x)=f(x+1)-f(x)$ on $x \in[0,0]$

then g is continuous on [0 , 1]

Also g(0) = f(1)-f(0)

Sol $\quad g(1)=-g(0)$

Thus by the IVT, $\exists x \in[0,1]$ s.t. $g(x)=0$

i.e $f(x+1)-f(x)=0$

i.e. $f(x+1)=f(x)$

Hence, done.