mathematics-class-12-unit-06-chapter-09-derivatives-l-9-12-62icb-zyqjq

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Local maxima and minima of f(x) </primary>
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- 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. $f(c) \leq f(x) \quad \forall x \in(c-h, c+h)$
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<footer style = "font-size: 18px"> $\rightarrow$ Derivatives lecture - 9 $\rightarrow$ <span style = "color:blue"> Local maxima and minima of f(x) </span> $\rightarrow$ Local maxima and minima of f(x) $\rightarrow$ Theorem </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Theorem </primary>
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- 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.
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<footer style = "font-size: 18px">Local maxima and minima of f(x) $\rightarrow$ Local maxima and minima of f(x) $\rightarrow$ <span style = "color:blue"> Theorem </span> $\rightarrow$ Theorem $\rightarrow$ Problem-1 </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Theorem </primary>
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- (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.
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<footer style = "font-size: 18px">Local maxima and minima of f(x) $\rightarrow$ Theorem $\rightarrow$ <span style = "color:blue"> Theorem </span> $\rightarrow$ Problem-1 $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- 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.
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<footer style = "font-size: 18px">Theorem $\rightarrow$ Theorem $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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-  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.
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<footer style = "font-size: 18px">Theorem $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $g(x) =2 x^3-6 x^2+6 x+5$

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

- $ =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.
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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Example $\rightarrow$ Second derivative test </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Example </primary>
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- 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

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

- In this example, the end derivative of the fx. is positive at the pt. of local min and  is negative at the point of local max.
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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Second derivative test $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Second derivative test </primary>
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- 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 $f^{\prime \prime}(c)=0$ )
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<footer style = "font-size: 18px">Solution $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Second derivative test </span> $\rightarrow$ Problem-3 $\rightarrow$ f''(c)=0 </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- 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.
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<footer style = "font-size: 18px">Example $\rightarrow$ Second derivative test $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ f''(c)=0 $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> f''(c)=0 </primary>
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- 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,
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<footer style = "font-size: 18px">Second derivative test $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> f''(c)=0 </span> $\rightarrow$ Example $\rightarrow$ Example </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Example </primary>
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- $ 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.

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<footer style = "font-size: 18px">Problem-3 $\rightarrow$ f''(c)=0 $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Example $\rightarrow$ Example  </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Example </primary>
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- 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.
- $f(0)=12$ ; $f(1)=7$ ; $f(-2)=-20$

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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Example </span> $\rightarrow$ Problem-4 $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Derivatives L-9 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- $f(x)=x+\frac{1}{x}, x \neq 0$

- 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
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<footer style = "font-size: 18px">Example $\rightarrow$ Example $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Thank you $\rightarrow$  </footer>