mathematics-class-12-unit-06-chapter-06-derivatives-l-6-12-1oe4eypi-fy

<h3 id="success-stylefont-size25px-derivatives-l-6-success"><Success style="font-size:25px"> Derivatives L-6 </Success></h3>
<h3 id="primary-stylefont-size24px--rolles-theorem-primary"><primary style="font-size:24px">  Rolle’s Theorem </primary></h3>
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<p>Let $f:[a, b] \rightarrow \mathbb{R}$ be a function satisfying the following:</p>
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<p>(i) $f$ is continuous on the closed interval $[a, b]$.</p>
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<p>(ii) $f$ is differentiable on the open interval $(a, b)$.</p>
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<p>(iii) $f(a)=f(b)$</p>
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<p>Then, there exists at least one $c \in(a, b)$ such that $f^{\prime}(c)=0$</p>
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<footer style = "font-size: 18px"> $\rightarrow$ Derivatives lecture-6 $\rightarrow$ <span style = "color:blue">  Rolle's Theorem </span> $\rightarrow$ Conditions for Rolle's theorem $\rightarrow$  Differentiability of functions </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px"> Conditions for Rolle's theorem </primary>
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- The assumptions in the theorem are necessary:

- (i) There is $no  c \in(a, b)$ such that $f^{\prime}(c)=0$.

- $f$ is differential in $(a, b)$

- $f(a)=f(b)$

- if is continuous at all points of $[a, b]$ except at a

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<footer style = "font-size: 18px">Derivatives lecture-6 $\rightarrow$  Rolle's Theorem $\rightarrow$ <span style = "color:blue"> Conditions for Rolle's theorem </span> $\rightarrow$  Differentiability of functions $\rightarrow$  Proof </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px">  Differentiability of functions </primary>
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- $\underline { \text {(ii) Differentiability of f on (a, b).}}$

- If in want on $[a, b]$

- $f$ in differentiable at all except one point in

- $(a, b)$
- $f(a)=f(b)$

- There is no $c$ such that $f^{\prime}(c)=0$.

- $ {\underline { (iii) \quad \text {f(a)=f(b)}}}$

- $ f(x)=x \text { on }[0,1]$     
- $ \text { if is cont on }[0,1] \quad\text { Here, } f(0) \neq f(1)$      
- $ \text { if is diff'ble on }(0,1) $     
- $ f^{\prime}(x)=1 \quad \forall x \in(0,1)$

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<footer style = "font-size: 18px"> Rolle's Theorem $\rightarrow$ Conditions for Rolle's theorem $\rightarrow$ <span style = "color:blue">  Differentiability of functions </span> $\rightarrow$  Proof $\rightarrow$ Example-1 </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px">  Proof </primary>
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- $\underline {\text{Idea of the proof:}}$

- Under the assumptions of the Rolle's theorem, we can show there is at least one point $c \in(a, b)$ where $f(x)$ attains its minimum or maximum value.

- Fact: Any continuous function $f:[a, b] \rightarrow \mathbb{R}$ is bounded, and $f$ attains its minimum and maximum values on $[a, b]$, ie., $\exists x_0, y_0 \in[a, b]$ s.t.
- $ f\left(x_0\right) \leqslant f(x) \leqslant f\left(y_0\right) \quad \forall x \in[a,b]$

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<footer style = "font-size: 18px">Conditions for Rolle's theorem $\rightarrow$  Differentiability of functions $\rightarrow$ <span style = "color:blue">  Proof </span> $\rightarrow$ Example-1 $\rightarrow$ Example-2 </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px"> Example-1 </primary>
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- Note that the previous result is not true for continuous functions on open interval $(a, b)$.

- For example, $f(x)=\tan x$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ is continuous but there is no minimum value nor maximum value

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<footer style = "font-size: 18px"> Differentiability of functions $\rightarrow$  Proof $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Example-2 $\rightarrow$ Proof of the Rolle's theorem </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px"> Example-2 </primary>
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- $ \underline{\text {Another fact:}}$ 
- If $f:[a, b] \rightarrow{\mathbb{R}}$ attains its minimum or maximum value in the open interval.
- (ie. not at the end points), then the derivative at that point must be zero if the function $f$ is differentiable at that point.

- e.g.

- The maximum value is attained at $x=\frac{1}{2}$ but the function is not differentiable at $x=\frac{1}{2}$

- $f'(\frac{1}{2})=0$
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<footer style = "font-size: 18px"> Proof $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Proof of the Rolle's theorem $\rightarrow$ Proof of the Rolle's theorem </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px"> Proof of the Rolle's theorem </primary>
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- Since $f$ in assumed to be continuous on the closed interval $[a, b]$, there exist
- $x_0, y_0 \in[a, b] \text { s.t. } f\left(x_0\right) \leq f(x) \leqslant f\left(y_0\right)$
- for all $x \in[a, b]$.

- $ {\underline {\text {Case I:}}}$    
- $x_0$ and $y_0$ are the end points $a, b$.

- But since $f(a)=f(b)$, we must have that $f(x)$ is a constant on $[a, b]$
- $\Rightarrow f^{\prime}(x)=0 \quad \forall x \in(a, b)$
- So, we can choose any $c \in(a, b)$ to get $f^{\prime}(c)=0$

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<footer style = "font-size: 18px">Example-1 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Proof of the Rolle's theorem </span> $\rightarrow$ Proof of the Rolle's theorem $\rightarrow$ Maximum or minimum  </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px"> Proof of the Rolle's theorem </primary>
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- $ \underline{\text{Case II:}}$ At least one of $x_0$ or $y_0$ lies in the open interval $(a, b)$.

- In this case, either the minimum on the maximum value of $f(x)$ is attained in the open interval $(a, b)$.

- Then $f^{\prime}$ must be zero at that point (because $f$ is assumed to be differentiable on open interval $(a, b))$.

- This proves the Rolle's theorm.

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<footer style = "font-size: 18px">Example-2 $\rightarrow$ Proof of the Rolle's theorem $\rightarrow$ <span style = "color:blue"> Proof of the Rolle's theorem </span> $\rightarrow$ Maximum or minimum  $\rightarrow$ Maximum or minimum  </footer>

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- Why $f^{\prime}$ must be zero at the point of maximum or minimum in an open interval?

- Suppose $f(c) \geqslant f(x) \quad \forall x \in[a, b]$ where $c \in(a, b)$

- Also, assume $f$ is differentially at $x=c$.

- $ \underline{\text{Claim:}}$ $f^{\prime}(c)=0$.
- suppose not Then either $f^{\prime}(c)>0$ or $f^{\prime}(c)<0$. If $f^{\prime}(c)>0$, then $\lim _{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}=f^{\prime}(c)>0$

- Then for small $h> 0, f(c+h)-f(c)>0$

- $\Rightarrow f(c+h)>f(c)$ for all small $h$.

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<footer style = "font-size: 18px">Proof of the Rolle's theorem $\rightarrow$ Proof of the Rolle's theorem $\rightarrow$ <span style = "color:blue"> Maximum or minimum  </span> $\rightarrow$ Maximum or minimum  $\rightarrow$ Mean Value Theorem </footer>

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### <primary style="font-size:24px"> Maximum or minimum  </primary>
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- But this contradicts $f(c)$ is the maximum value Similarly, we get a contradiction if $f^{\prime}(c)<0$

- Hence, $f^{\prime}(c)=0$.

- Why $f^{\prime}$ must be zero at the point of maximum or minimum in an open interval?

- Suppose $f(c) \geqslant f(x) \quad \forall x \in[a, b]$ where $c \in(a, b)$

- Also, assume $f$ is differentially at $x=c$.

- Then for small $h> 0, f(c+h)-f(c)>0$

- $\Rightarrow f(c+h)>f(c)$ for all small $h$.

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<footer style = "font-size: 18px">Proof of the Rolle's theorem $\rightarrow$ Maximum or minimum  $\rightarrow$ <span style = "color:blue"> Maximum or minimum  </span> $\rightarrow$ Mean Value Theorem $\rightarrow$ Proof </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px"> Mean Value Theorem </primary>
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- Let $f:[a, b] \rightarrow \mathbb{R}$ be such that

- (i) $f(x)$ is continuous on the closed interval $[a,b]$

- (ii) $f(x)$ is differentiable on the open interval $(a, b)$

- Then there exists at least one $c \in(a, b)$

- such that 
- $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$

- Note that this is a generalization of the Rolle's theorem as if $f(a)=f(b)$ then

- $M V T$ $\Rightarrow \exists c \in(a, b)$ st.

- $f^{\prime}(c)=0$ ie; Rolle's theorem

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<footer style = "font-size: 18px">Maximum or minimum  $\rightarrow$ Maximum or minimum  $\rightarrow$ <span style = "color:blue"> Mean Value Theorem </span> $\rightarrow$ Proof $\rightarrow$ Proof </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px"> Proof </primary>
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- $ \underline {\text{Proof:}}$

- Slope of the secant line joinining $(a, f(a) ~ and ~ (b, f(b)$

- is $\frac{f(b)-f(a)}{b-a}=m$

- We need to show that $\exists c \in(a, b)$ s.t.

- the slope of the tangent line at $(c, f(c))$ is the same as $m$.

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<footer style = "font-size: 18px">Maximum or minimum  $\rightarrow$ Mean Value Theorem $\rightarrow$ <span style = "color:blue"> Proof </span> $\rightarrow$ Proof $\rightarrow$ Proof </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px"> Proof </primary>
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- The eqn. of the line joining $(a, f(a)),(b, f(b))$ is given by

- $y-f(a)=\frac{f(b)-f(a)}{b-a}(x-a)$

- Let  L(x)=f(a)+ $ [\frac{f(b)-f(a)}{b-a}].(x-a)$

- Let $g(x)=f(x)-L(x)$

- Then $g$ is continuous on $[a, b]$;

- $g$ is differentiable on $(a, b)$.

- Also,

- $ g(a) =f(a)-L(a)=f(a)-f(a)=0$    
- $ g(b) =f(b)-L(b) \quad \therefore g(a)=g(b)$     
- $ =f(b)-f(b)=0 \quad $

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<footer style = "font-size: 18px">Mean Value Theorem $\rightarrow$ Proof $\rightarrow$ <span style = "color:blue"> Proof </span> $\rightarrow$ Proof $\rightarrow$ Applications of the MVT </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
### <primary style="font-size:24px"> Proof </primary>
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- So, by the Rolle's theorem, $\exists c \in(a, b)$ such that
- $g^{\prime}(c)=0$

- But $g^{\prime}(x)=f^{\prime}(x)-L^{\prime}(x)$

- $ =f^{\prime}(x)-\left[\frac{f(b)-f(a)}{b-a}\right],$

- $ \therefore g^{\prime}(c)=0  \Rightarrow f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$

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<footer style = "font-size: 18px">Proof $\rightarrow$ Proof $\rightarrow$ <span style = "color:blue"> Proof </span> $\rightarrow$ Applications of the MVT $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Derivatives L-6 </Success>
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- $\underline {\text{Corollary 1:}}$   Suppose $f:[a, b] \rightarrow \mathbb{R}$ be continuous and $f^{\prime}(x)=0$ for all $x \in(a, b)$.

- Then $f$ must be a constant.

- $\underline {\text{Proof:}}$

- Let $a \leq x_1<x_2 \leq b$. We have to show that $f\left(x_1\right)=f\left(x_2\right)$.

- By $M V T, \exists c \in\left(x_1, x_2\right)$ set.

- $f^{\prime}(c)=\frac{f\left(x_2\right)-f\left(x_1\right)}{x_2-x_1}$

- But $f^{\prime}(c)=0$. Therefore, $f\left(x_2\right)=f\left(x_1\right)$.

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<footer style = "font-size: 18px">Proof $\rightarrow$ Proof $\rightarrow$ <span style = "color:blue"> Applications of the MVT </span> $\rightarrow$ Thank you $\rightarrow$  </footer>