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Question: How do you find the value of c guaranteed by the mean value theorem for \[f(x) = \dfrac{{2x}}{{{x^2}...

How do you find the value of c guaranteed by the mean value theorem for f(x)=2xx2+1f(x) = \dfrac{{2x}}{{{x^2} + 1}} on the interval [0,1][0,1] .

Explanation

Solution

Hint : Here in this question we have to find the value of c, such that the given function will obey the mean value theorem. By using the definition of mean value theorem we have to determine the value of c and the value of c will lies in the given interval. hence we obtain the required solution for the given question.

Complete step by step solution:
The mean value theorem states that the function ff is continuous on [a,b][a,b] and differentiable on (a,b)(a,b) then there exist a point c such that f′(c)=f(b)−f(a)b−af'(c) = \dfrac{{f(b) - f(a)}}{{b - a}}
Here the given function is f(x)=2xx2+1f(x) = \dfrac{{2x}}{{{x^2} + 1}} on the interval [0,1][0,1] . This function will satisfy the mean value theorem.
Therefore, we have f′(c)=f(b)−f(a)b−af'(c) = \dfrac{{f(b) - f(a)}}{{b - a}} , here the value of b is 1 and the value of a is 0.
Therefore we have
⇒f′(c)=f(1)−f(0)1−0\Rightarrow f'(c) = \dfrac{{f(1) - f(0)}}{{1 - 0}}
The f(1)=2(1)12+1f(1) = \dfrac{{2(1)}}{{{1^2} + 1}}
On simplifying we get
f(1)=22=1f(1) = \dfrac{2}{2} = 1
The f(0)=2(0)02+1f(0) = \dfrac{{2(0)}}{{{0^2} + 1}}
On simplifying we get
f(0)=0f(0) = 0
Now the f′(c)f'(c) is written as
⇒f′(c)=1−01\Rightarrow f'(c) = \dfrac{{1 - 0}}{1}
On simplifying we get
⇒f′(c)=1\Rightarrow f'(c) = 1
now consider the given function
f(x)=2xx2+1f(x) = \dfrac{{2x}}{{{x^2} + 1}}
On differentiating the function with respect x we get
f′(x)=(x2+1)2−2x(2x)(x2+1)2f'(x) = \dfrac{{({x^2} + 1)2 - 2x(2x)}}{{{{\left( {{x^2} + 1} \right)}^2}}}
on simplifying
⇒f′(x)=2x2+2−4x2(x2+1)2\Rightarrow f'(x) = \dfrac{{2{x^2} + 2 - 4{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}
⇒f′(x)=2−2x2(x2+1)2\Rightarrow f'(x) = \dfrac{{2 - 2{x^2}}}{{{{\left( {{x^2} + 1} \right)}^2}}}
Now replace x by c we get
⇒f′(c)=2−2c2(c2+1)2\Rightarrow f'(c) = \dfrac{{2 - 2{c^2}}}{{{{\left( {{c^2} + 1} \right)}^2}}}
As we know that the value of f′(c)=1f'(c) = 1 , hence the above equation is written as
⇒1=2−2c2(c2+1)2\Rightarrow 1 = \dfrac{{2 - 2{c^2}}}{{{{\left( {{c^2} + 1} \right)}^2}}}
Take (c2+1)2{\left( {{c^2} + 1} \right)^2} to LHS and the equation is rewritten as
⇒(c2+1)2=2−2c2\Rightarrow {\left( {{c^2} + 1} \right)^2} = 2 - 2{c^2}
The algebraic formula (a+b)2=a2+b2−2ab{\left( {a + b} \right)^2} = {a^2} + {b^2} - 2ab , considering the formula and the equation is written as
⇒c4+1+2c2=2−2c2\Rightarrow {c^4} + 1 + 2{c^2} = 2 - 2{c^2}
Rearranging the equation it is written as
⇒c4+4c2−1=0\Rightarrow {c^4} + 4{c^2} - 1 = 0
This is the quadratic equation we determine the solution for c2{c^2}
therefore we have
⇒c2=−(4)±42−4(1)(−1)2(1)\Rightarrow {c^2} = \dfrac{{ - (4) \pm \sqrt {{4^2} - 4(1)( - 1)} }}{{2(1)}}
⇒c2=−4±16+42\Rightarrow {c^2} = \dfrac{{ - 4 \pm \sqrt {16 + 4} }}{2}
⇒c2=4±202\Rightarrow {c^2} = \dfrac{{4 \pm \sqrt {20} }}{2}
On further simplification
⇒c2=2±5\Rightarrow {c^2} = 2 \pm \sqrt 5
Taking the square root on both sides we get
⇒c=±2±5\Rightarrow c = \pm \sqrt {2 \pm \sqrt 5 }
Therefore the value of c is ±2±5\pm \sqrt {2 \pm \sqrt 5 }
Hence we have obtained the solution for the given question.
So, the correct answer is “ ±2±5\pm \sqrt {2 \pm \sqrt 5 } ”.

Note : The question is related to the mean value theorem. The theorem states that if the function is continuous on [a,b][a,b] and differentiable on (a,b)(a,b) then there exist a point c such that f′(c)=f(b)−f(a)b−af'(c) = \dfrac{{f(b) - f(a)}}{{b - a}} . While simplifying we get the equation in the form of quadratic equation we must know the formula x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}