mathematics-class-12-unit-06-chapter-08-derivatives-l-8-12-qxrl8wuumg4

### <Success style="font-size:25px"> Derivatives L-8 </Success>
### <primary style="font-size:24px"> Recall </primary>
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- Finding the maximum and minimum values of a given function $f(x)$ on a given interval I.

- <ins>**Recall:**</ins>      
- <ins>**Theorem:**</ins> If $f(x)$ is a continuous function on a closed interval $[a, b]$, then the function attains its maximum value and minimum on the interval $[a, b]$ ie. Here exist points $x_0$ and $y_0 \in[a, b]$

- 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"> $\rightarrow$ Derivatives lecture - 8 $\rightarrow$ <span style = "color:blue"> Recall </span> $\rightarrow$ minimum and maximum value $\rightarrow$ Problem-1 </footer>

### <Success style="font-size:25px"> Derivatives L-8 </Success>
### <primary style="font-size:24px"> minimum and maximum value </primary>
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- In this case $f\left(x_0\right)$ is the minimum value of $f(x)$ on the interval $[a, b]$ and it is attained at the point $x_0 \in[a, b]$.

- Similarly, $f\left(y_0\right)$ is the maximum value of $f(x)$ on the interval $(a, b)$ and it is attained at the point $y_0 \in[a, b]$.

- Note that there can be more than one point at which the minimum (or maximum) value is attained.

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<footer style = "font-size: 18px">Derivatives lecture - 8 $\rightarrow$ Recall $\rightarrow$ <span style = "color:blue"> minimum and maximum value </span> $\rightarrow$ Problem-1 $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Derivatives L-8 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- How to find the minimum and maximum value ?

- <ins>eg. 1</ins>    
- The minimum value \& the maximum can be attained at points in the open interval $(a, b)$.   
- In this example, $f^{\prime}\left(x_0\right)=0$ \& $f^{\prime}\left(y_0\right)=0$.

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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Conclusion $\rightarrow$ Definition </footer>

### <Success style="font-size:25px"> Derivatives L-8 </Success>
### <primary style="font-size:24px"> Conclusion </primary>
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- **Conclusion:** The points of absolute minima and maxima of $f(x)$ on $[a, b]$ can be either one of the end points or an interior point.

- <ins>**Recall:**</ins>

- <ins>**Theorem:-**</ins>If $f(x)$ has minima or maxima on the open interval $(a, b)$, then either the derivative does not exist at that point or the derivative $f^{\prime}(x)=0$ at that point.

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Conclusion </span> $\rightarrow$ Definition $\rightarrow$ Steps to find maxima or minima </footer>

### <Success style="font-size:25px"> Derivatives L-8 </Success>
### <primary style="font-size:24px"> Steps to find maxima or minima </primary>
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- <ins>**Step 1:-**</ins> Find all critical points in the open interval $(a, b)$, ie, the points where $f^{\prime}(x)=0$ or $f^{\prime}(x)$ does not exist.

- <ins>**Step 2:-**</ins> Find the values of $f(x)$ at all critical points.

- <ins>**Step 3:-**</ins> Find the values of $f(x)$ at the end points, ie., find $f(a)$ \& $f(b)$.

- <ins>**Step 4:-**</ins> Find the minimum and the maximum values among those obtained in steps $2$ \& $3$.

- This gives us the minimum \& the maximum values and also the points of global (absolute) minima \& maxima.

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<footer style = "font-size: 18px">Conclusion $\rightarrow$ Definition $\rightarrow$ <span style = "color:blue"> Steps to find maxima or minima </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

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### <primary style="font-size:24px"> Problem-2 </primary>
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- <ins>**Example:-**</ins> Find the maximum and minimum values of $f(x)=x^3-6 x^2+9 x+15$ on $[-1,2]$    
- Find the critical points.     
- Since $f(x)$ is a polynomial, $f^{\prime}(x)$ exists at all points.     
- $f^{\prime}(x)=3 x^2-12 x+9$

- Find the zeros of $f^{\prime}(x)$ :

- $ 3 x^2-12 x+9=0 $   
- $ \Leftrightarrow  x^2-4 x+3=0 $   
- $ \Leftrightarrow  (x-1)(x-3)=0 $   
- $ \Leftrightarrow  x=1 \text { or } x=3 .$

- Now, $x=1$ lies in the interval $(-1,2)$ where   
- $x=3$ lies outside the interval.

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<footer style = "font-size: 18px">Definition $\rightarrow$ Steps to find maxima or minima $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Derivatives L-8 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- <ins>**Example:-**</ins> Find the maximum and minimum values of $f(x)=x^3-6 x^2+9 x+15$ on $[-1,2]$    
- Find th critical points.     
- Since $f(x)$ is a polynomial, $f^{\prime}(x)$ exists at all points.     
- $f^{\prime}(x)=3 x^2-12 x+9 $
- Find the zeros of $f^{\prime}(x)$ :

- $ 3 x^2-12 x+9=0 $   
- $ \Leftrightarrow  x^2-4 x+3=0 $   
- $ \Leftrightarrow  (x-1)(x-3)=0 $   
- $ \Leftrightarrow  x=1 \text { or } x=3 .$

- Now, $x=1$ lies in the interval $(-1,2)$ where   
- $x=3$ lies outside the interval.

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<footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem 3 </span> $\rightarrow$ Remark $\rightarrow$ Remark </footer>

### <Success style="font-size:25px"> Derivatives L-8 </Success>
### <primary style="font-size:24px"> Remark </primary>
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- Consider $f(x)=x+\frac{1}{x}, x \in \mathbb{R} \backslash ${0}$ $.

- Find the minimum and maximum values of $f(x)$ if they exist.

- <ins>**Remark:-**</ins> Since here we are not considering $f(x)$ over a closed interval, the min. value or the max. value need not exist.

- $ f(x)=x+\frac{1}{x}=\frac{x^2+1}{x}, x \neq 0$

- Note that $x^2+1-2 x=(x-1)^2 \geqslant 0$

- $ \Rightarrow x^2+1 \geqslant 2 x $   
- $ \Rightarrow \frac{x^2+1}{x} \geqslant 2 $

- if $x>0$ \& $\frac{x^2+1}{x}<0$ if $x=0 $

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem 3 $\rightarrow$ <span style = "color:blue"> Remark </span> $\rightarrow$ Remark $\rightarrow$ Problem 4 </footer>

### <Success style="font-size:25px"> Derivatives L-8 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
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- Is there any maximum value on $(0, \infty)$ ?

- $ f(x) \geqslant x+ \frac{1}{2} $   
- $\rightarrow +\infty $ as $ x \rightarrow 0+ $   
- $ \rightarrow +\infty $ as $ x \rightarrow +\infty $

- $\therefore f(x)$ has no maximum value on $(0, \infty)$.    
- Now on $(-\infty, 0), f(x)=x+\frac{1}{x}$   
- $ =-\left(-x+\frac{1}{-x}\right)$
- $....-x>0$

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<footer style = "font-size: 18px">Remark $\rightarrow$ Remark $\rightarrow$ <span style = "color:blue"> Problem 4 </span> $\rightarrow$ Solution $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Derivatives L-8 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $\therefore$ On $(-\infty, 0),-f(x)=-x+\frac{1}{-x} \geqslant 2$ (by previous part)
- $ \therefore f(x) \leqslant-2 \text { on }(-\infty, 0)$

- Also, as $x \rightarrow-\infty, f(x)=x+\frac{1}{x} \rightarrow-\infty$

- \& as $x \rightarrow 0-, f(x) \rightarrow-\infty$.

- So, $f(x)$ has max value -2 obtained at $x=-1$

- and it has no min. value on $(-\infty, 0)$.

- <ins>**Another way:-**</ins> For  $x<0$,

- $ x^2+1+2 x=(x+1)^2 \geqslant 0 $   
- $ \Rightarrow x^2+1 \geqslant-2 x $   
- $ \Rightarrow \frac{x^2+1}{-x} \geqslant 2 \quad (\because-x>0) $   
- $ \Rightarrow \frac{x^2+1}{x} \leqslant-2 $

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<footer style = "font-size: 18px">Remark $\rightarrow$ RProblem 4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Derivatives L-8 </Success>
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-  Can we use Calculus to derive the same conclusion ?

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

- $ f(x)$ is continuous \& differentiable for all $x \neq 0$.   
- $ f^{\prime}(x)=1-\frac{1}{x^2} $   
- $ f^{\prime}(x)=0 \Leftrightarrow x^2=1 \Leftrightarrow x= \pm 1 $.

- Also, $x^2>1 \Rightarrow f^{\prime}(x)=1-\frac{1}{x^2}>0 $   
- if $x^2<1 \Rightarrow \frac{1}{x^2}>1 \Rightarrow 1-\frac{1}{x^2}<0$   
- $x^2<1 \Leftrightarrow-1<x<1$    
- $x^2>1 \Leftrightarrow x>1$ or $x<-1$

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<footer style = "font-size: 18px">Problem 4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Solution $\rightarrow$ Remark </footer>

### <Success style="font-size:25px"> Derivatives L-8 </Success>
### <primary style="font-size:24px"> Remark </primary>
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-  On $\mathbb{R} \backslash ${0}$, f(x)=x+\frac{1}{x}$ has no minimum value and no maximum value.

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<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>