What does the mean value theorem mean?
Answer
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Hint: Mean value theorem is the part of differentiation concept in calculus. This is a very important result. You have to know differentiation before knowing this. This result is very important and it has many important applications.
Complete step by step solution:
According to the mean value theorem, for a function f which is continuous in an closed interval and also differentiable in an open interval, then there will always be a value c which exits in the interval such that the differentiation of f at c that is f'(c) is always equal to the average rate of change in the interval.
For example, we will take a question. How do you determine all values of c that satisfy the mean value theorem on the interval [0, 1] for $ f\left( x \right)=\dfrac{x}{x+6}$ . This can be solved by doing,
$ f\left( x \right)=\dfrac{x}{x+6}$--- ( 1 )
The derivative of f(x) :
$ \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{6}{{{\left( x+6 \right)}^{2}}}$ ------(2)
According to mean value theorem, we get
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{f\left( 1 \right)-f\left( 0 \right)}{1-0} $------(3)
$ \Rightarrow f\left( 1 \right)=\dfrac{1}{1+6}=\dfrac{1}{7}$ ------ (4)
$ \Rightarrow f\left( 0 \right)=\dfrac{0}{0+6}=0$ ------(5)
We have to substitute 4 and 5. After substituting, we get:
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{\dfrac{1}{7}-0}{1-0}=\dfrac{1}{7}$
We have to equate 2 and 5. Therefore, we get:
$ \Rightarrow \dfrac{6}{{{\left( x+6 \right)}^{2}}}=\dfrac{1}{7}$
$ \Rightarrow {\left( x+6 \right)}^{2}=42$
$ \Rightarrow { x+6}=\pm{6.480741}$
$ \Rightarrow { x+6}=+{6.480741} \Rightarrow { x}={0.480741}$
$ \Rightarrow { x+6}=-{6.480741} \Rightarrow { x}={-12.480741 }$
The answers are x = 0.480741 and x = -12.480741. Only x = 0.480741 can be found in the range [0,1]. So, it is the only c that satisfies the mean value theorem.
Note: The theorem can be understood by doing some examples. So, it is important that you do some problems relating to this theorem. You have to do the sums step by step like shown in the above example so that you can understand it easily.
Complete step by step solution:
According to the mean value theorem, for a function f which is continuous in an closed interval and also differentiable in an open interval, then there will always be a value c which exits in the interval such that the differentiation of f at c that is f'(c) is always equal to the average rate of change in the interval.
For example, we will take a question. How do you determine all values of c that satisfy the mean value theorem on the interval [0, 1] for $ f\left( x \right)=\dfrac{x}{x+6}$ . This can be solved by doing,
$ f\left( x \right)=\dfrac{x}{x+6}$--- ( 1 )
The derivative of f(x) :
$ \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{6}{{{\left( x+6 \right)}^{2}}}$ ------(2)
According to mean value theorem, we get
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{f\left( 1 \right)-f\left( 0 \right)}{1-0} $------(3)
$ \Rightarrow f\left( 1 \right)=\dfrac{1}{1+6}=\dfrac{1}{7}$ ------ (4)
$ \Rightarrow f\left( 0 \right)=\dfrac{0}{0+6}=0$ ------(5)
We have to substitute 4 and 5. After substituting, we get:
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{\dfrac{1}{7}-0}{1-0}=\dfrac{1}{7}$
We have to equate 2 and 5. Therefore, we get:
$ \Rightarrow \dfrac{6}{{{\left( x+6 \right)}^{2}}}=\dfrac{1}{7}$
$ \Rightarrow {\left( x+6 \right)}^{2}=42$
$ \Rightarrow { x+6}=\pm{6.480741}$
$ \Rightarrow { x+6}=+{6.480741} \Rightarrow { x}={0.480741}$
$ \Rightarrow { x+6}=-{6.480741} \Rightarrow { x}={-12.480741 }$
The answers are x = 0.480741 and x = -12.480741. Only x = 0.480741 can be found in the range [0,1]. So, it is the only c that satisfies the mean value theorem.
Note: The theorem can be understood by doing some examples. So, it is important that you do some problems relating to this theorem. You have to do the sums step by step like shown in the above example so that you can understand it easily.
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