Def: A pt. c in the domain of $f(x)$ is called a pt. of local maxima if there exist $h>0$ such $f(c) \geqslant f(x) \quad \forall x \in(c-h, c+h)$

(ie. $f(c)$ is the max. value of $f(x)$ is some small enough interval $(c-h, c+h)$ containing the pt.c )

similarly :

$c$ is called a pt. of local minima if $\exists h>0$ sit. $f(c) \leq f(x) \quad \forall x \in(c-h, c+h)$

$c_1$ is a pt. of local maxima.

$c_2$.. .. . . . minima.

$C_3$ is ". ". .. "I maxima.

$c_4$. . . .. . minima.

Theorem: If $c$ is a point of local maxima or local minima, then either $f^{\prime}(c)=0$ or $f^{\prime}(c)$ does not exist.

Theorem (First Derivative Test):

Let $f(x)$ be a function defined on an open interval I.

(i) If $f^{\prime}(x)$ changes sign from +ve to -ve, so we move across c, then c is a pt. of local minima.

(ii) If $f^{\prime}(x)$ changes sign from - ve to +ve as we move across c, than c is a pt. of local minima.

(iii) If $f^{\prime}(x)$ does not change sign as we move across $c$, then $c$ is neither a pt. of local max. nor a pt. of local min.

So, to find pts. of local min. R local max. we find the critical pts (ie. it where $f^{\prime}(x)=0$ or $f^{\prime}(x)$ does not exist)

and then we can use the first derivative test to determine whether those pt. are points of local max, local min. or neither.

eg. $f(x)=x^3$ on $\mathbb{R}$.

$f^{\prime}(x)=3 x^2$

Thus, $f^{\prime}(x)=0 \Leftrightarrow x=0$ ie. 0 is th a on 6 critical

Since $f^{\prime}(x)=3 x^2>0 \quad \forall x>0$

$f^{\prime}(x)$ does not change sign do we move actors $x=0$

sq $x=0$ is neither a pt. of local max. nor a pt. of local min.

Such a pt. which is a critical pt. but is neither a pt. of local max nor local min is called an inflection point.

Find the pts. of local maxima and local minima of

$\begin{aligned}& f(x)=x^3-3 x+3 \\& f^{\prime}(x)=3 x^2-3=3\left(x^2-1\right)=3(x-1)(x+1) \\& f^{\prime}(x)=0 \Leftrightarrow x=-1 \text { or } x=1\end{aligned}$

Thus, $x=-1$ is a pt. of local maxima & $x=1$ is a pt. of local minima.

$g(x) =2 x^3-6 x^2+6 x+5$

$g^{\prime}(x) =6 x^2-12 x+6$

$=6\left(x^2-2 x+1\right)=6(x-1)^2$

$x=1$ is a critical pt.

But $g^{\prime}(x)>0$ as we move across $x=1$

therefore $x=1$ is a point of inflection for $g(x)$.

(neither a local maxima nor a local minima) $g(x)$ has no local max. or local min.

Let's look at $f(x)=x^2$

$f^{\prime}(x)=2 x \quad \text { So, }$ $x=0$ is the only critical point.

By the first derivative test to see that $x=0$ is a pt. of local minima.

For $g(x)=-x^2, x=0$ is a $p t$. of local maxima

$f^{\prime \prime}(x)=2$, $g^{\prime \prime}(x)=-2$

Theorem: Suppose $f(x)$ is a fraction which is twice differentiable on an open interval I. Also, suppose $f^{\prime}(c)=0$. Then

(i) If $f^{\prime \prime}(c)>0$ then $c$ is a pt. of local min.

(ii) If $f^{\prime \prime}(c)<0$ then $c$ is a pt. of local max.

(iii) If $f^{\prime \prime}(c)=0$ then the test fails (ie. we cannot conclude anything if $f^{\prime \prime}(c)=0$ )

Consider $f(x)=x^4, g(x)=-x^4$

Then $f^{\prime}(0)=0=g^{\prime}(0)$

Also, $f^{\prime \prime}(0)=0=g^{\prime \prime}(0)$

We can see by the first derivative test or by direct observation that $f(x)$ has a local min. at $x=0$ whereas $g(x)$ has a local max. at $x=0$.

consider $h(x)=x^3, h^{\prime}(x)=3 x^2, h^{\prime \prime}(x)=6 x$

$h^{\prime}(0)=0, h^{\prime \prime}(0)=0$

Here $x=0$ ' is neither a local max. nor a local min.

So, if the second derivative $f^{\prime \prime}(c)=0$ at a critical pt. c, then we can have all possible cases. In such case we may try to use the first derivative test.

Proof of second derivative test:

(i) Suppose $f^{\prime}(c)=0$ and $f^{\prime \prime}(c)<0$

Claim: $c$ is a pt. of local minima.

for this we need to find

$h>0$ such that $f(c) \leqslant f(x) \quad \forall x \in(c-h, c+h)$.

By the definition of derivative,

$f^{\prime \prime}(c)=\lim _{x \rightarrow c}$

$\frac{f^{\prime}(x)-f^{\prime}(c)}{x-c} \cdot \text { But } f^{\prime}(c)=0$

$\therefore \lim _{x \rightarrow c} \frac{f^{\prime}(x)}{x-c}=f^{\prime \prime}(c) > 0$

$\Rightarrow \exists h>0 \text { set. } \frac{f^{\prime}(x)}{x-c}>0$

for all $x \in(c-h, c+h), x \neq c$

$\Rightarrow$ if $x \in(c, c+h)$ then $f^{\prime}(x)>0$

$[\therefore \text {Denominator} ~ x-c > 0]$

$\therefore$ By the list derivative test, $x=c$ is a pt. of local minima.

(ii) If $f^{\prime}(c)=0$ and $f^{\prime \prime}(c)<0$, then $x = c$ is a pt. of local max. can be proved in a similar way.

Find the pts. of local min. and local max of

$f(x)=3 x^4+4 x^3-12 x^2+12 .$

$f^{\prime}(x)=12 x^3+12 x^2-24 x$

$f^{\prime \prime}(x)=36 x^2+24 x-24$

Critical pts: $f^{\prime}(x)=0$ $\Leftrightarrow$ $12 x\left(x^2+x-2\right)=0$

$\Leftrightarrow 12 x(x-1)(x+2)=0$

$\Leftrightarrow$ $x=0$ or 1 or -2

$f^{\prime \prime}(x)=36 x^2+24 x-24$

Now, $\quad f^{\prime \prime}(0)=-24<0 \Rightarrow x=0$ is a pt. of local max.

$f^{\prime \prime}(1)=36+24-24=36>0$

$\Rightarrow x=1$ is a pt. of local min.

$f^{\prime \prime}(-2) =12\left[3(-2)^2+2(-2)-2\right]$

$=12[12-4-2]>0$

$\Rightarrow x=-2$ is a pt. of local min.

So, $x=0$ is a point of local max. $x=-2 \quad \& \quad x=1$ are pt. of local min.

$f(0)=12$ ; $f(1)=7$ ; $f(-2)=-20$

$f(x)=x+\frac{1}{x}, x \neq 0$

Find pts. of local min. and local max.

$f^{\prime}(x)=1-\frac{1}{x^2} \Rightarrow x= \pm 1 \text { are critical points }$

$f^{\prime \prime}(x)=\frac{2}{x^2}$

$f^{\prime \prime}(1)=2>0 \Rightarrow x=1$ is a pt. of local minima

$f^{\prime \prime}(-1)=-2<0 \Rightarrow x=-1$ is a point. of local maxima