Question

Verifiy rolles theorem for f(x)=log(x^2 + 2) - log^3 in [ -1,1]

Answer 1

$\large\underline\purple{\bold{Solution :- }}$

$\tt \:  f(x) = log( {x}^{2} + 2 ) - log(3) \: in \: [- 1, 1]$

Step :- 1

$\tt \:   \longrightarrow \:as \: x \: \epsilon \: [- 1, 1] \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt\implies \: \: \tt \:  {x}^{2} + 2 > 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \tt\implies \: log( {x}^{2} + 2 ) \: is \: defined$

$\tt\implies \:f(x) \: is \: continuous \: in \: [- 1, 1]$

Step :- 2

Differentiate f(x) w. r. t. x, we get

$\tt \:   \longrightarrow \:f'(x) \: = \dfrac{1}{ {x}^{2} + 2 } (2x) - 0$

$\tt \:   \longrightarrow \:f'(x) = \dfrac{2x}{ {x}^{2} + 2}$

$\tt\implies \:f(x) \: is \: differentiable \: in \: (- 1, 1)$

Step :- 3

$\tt \:   \longrightarrow \: f( - 1) = log(1 + 2) - log(3) = 0 \\ \tt \:   \longrightarrow \:f(1) = log(1 + 2) - log(3) = 0 \: \: \: \\ \tt\implies \: \: f( - 1) = f(1) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

☆ This implies, Rolle's Theorem is applicable. Therefore, there exist atleast one real number 'c' belongs to (- 1, 1) such that f'(c) = 0.

$\tt\implies \:\dfrac{2c}{ {c}^{2} + 2 } = 0 \: \: \: \: \: \: \: \: \\ \bf\implies \:c \: = 0 \: \epsilon \: (- 1, 1)$