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\[f(x) = {\tan ^{ - 1}}x\] on [0, 1]

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Hint: We need to verify the given function using the conditions for Lagrange’s mean value theorem.It states that function f(x) should be continuous in the closed interval [a, b] and differentiable on the open interval (a, b). If f (a) = f (b) then there exists at least one value of x between a, b and let that value be c, such that f’(c) =0 and verify the theorem.

Complete step-by-step answer:

Given function is \[f(x) = {\tan ^{ - 1}}x\] on interval [0, 1]

And \[{\tan ^{ - 1}}x\] has a unique value for all $x$ between 0 and 1.

$\therefore f(x)$ is continuous in [0,1]

$f(x) = {\tan ^{ - 1}}x$

Differentiating $f(x)$ with respect to $x$

$f'(x) = \dfrac{1}{{1 + {x^2}}}$

The value of ${x^2}$ is greater than 0.

$ \Rightarrow 1 + {x^2} > 0$

$\therefore f(x)$ is differentiable in (0,1)

So both necessary conditions of Lagrange’s mean value theorem are satisfied.

$\therefore $There exists a point $c \in (0,1)$ such that:

$f'(c) = \dfrac{{f(1) - f(0)}}{{1 - 0}}$

$ \Rightarrow f'(c) = f(1) - f(0)$

We have \[f'(x) = \dfrac{1}{{1 + {x^2}}}\]

\[ \Rightarrow f'(c) = \dfrac{1}{{1 + {c^2}}}\]

For $f(1) = {\tan ^{ - 1}}1 = \dfrac{\pi }{4}$

For $f(0) = {\tan ^{ - 1}}0 = 0$

We have $f'(c) = f(1) - f(0)$

$

\Rightarrow \dfrac{1}{{1 + {c^2}}} = \dfrac{\pi }{4} - 0 \\

\Rightarrow \dfrac{1}{{1 + {c^2}}} = \dfrac{\pi }{4} \\

\Rightarrow 4 = \pi \left( {1 + {c^2}} \right) \\

\Rightarrow \dfrac{4}{\pi } = 1 + {c^2} \\

\Rightarrow {c^2} = \dfrac{4}{\pi } - 1 \\

\Rightarrow c = \sqrt {\dfrac{4}{\pi } - 1} \simeq 0.52 \in (0,1) \\

$

Hence, Lagrange’s mean value theorem is verified.

Note: Lagrange’s mean value theorem is the mean value theorem itself or also called first mean value theorem.It states that if a function f(x) is continuous on a closed interval [a, b] then there is at least one point c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to the tangent at c.The function f(x) should be continuous in the closed interval [a, b] and differentiable on the open interval (a, b). If f (a) = f (b) then there exists at least one value of x between a, b and let that value be c, such that f’(c) =0.Students should remember the derivatives of trigonometric and inverse trigonometric functions.

Complete step-by-step answer:

Given function is \[f(x) = {\tan ^{ - 1}}x\] on interval [0, 1]

And \[{\tan ^{ - 1}}x\] has a unique value for all $x$ between 0 and 1.

$\therefore f(x)$ is continuous in [0,1]

$f(x) = {\tan ^{ - 1}}x$

Differentiating $f(x)$ with respect to $x$

$f'(x) = \dfrac{1}{{1 + {x^2}}}$

The value of ${x^2}$ is greater than 0.

$ \Rightarrow 1 + {x^2} > 0$

$\therefore f(x)$ is differentiable in (0,1)

So both necessary conditions of Lagrange’s mean value theorem are satisfied.

$\therefore $There exists a point $c \in (0,1)$ such that:

$f'(c) = \dfrac{{f(1) - f(0)}}{{1 - 0}}$

$ \Rightarrow f'(c) = f(1) - f(0)$

We have \[f'(x) = \dfrac{1}{{1 + {x^2}}}\]

\[ \Rightarrow f'(c) = \dfrac{1}{{1 + {c^2}}}\]

For $f(1) = {\tan ^{ - 1}}1 = \dfrac{\pi }{4}$

For $f(0) = {\tan ^{ - 1}}0 = 0$

We have $f'(c) = f(1) - f(0)$

$

\Rightarrow \dfrac{1}{{1 + {c^2}}} = \dfrac{\pi }{4} - 0 \\

\Rightarrow \dfrac{1}{{1 + {c^2}}} = \dfrac{\pi }{4} \\

\Rightarrow 4 = \pi \left( {1 + {c^2}} \right) \\

\Rightarrow \dfrac{4}{\pi } = 1 + {c^2} \\

\Rightarrow {c^2} = \dfrac{4}{\pi } - 1 \\

\Rightarrow c = \sqrt {\dfrac{4}{\pi } - 1} \simeq 0.52 \in (0,1) \\

$

Hence, Lagrange’s mean value theorem is verified.

Note: Lagrange’s mean value theorem is the mean value theorem itself or also called first mean value theorem.It states that if a function f(x) is continuous on a closed interval [a, b] then there is at least one point c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to the tangent at c.The function f(x) should be continuous in the closed interval [a, b] and differentiable on the open interval (a, b). If f (a) = f (b) then there exists at least one value of x between a, b and let that value be c, such that f’(c) =0.Students should remember the derivatives of trigonometric and inverse trigonometric functions.