Verify Roll�s Theorem for the function :f(x) = log (x² + 2) - log 3 in [-1, 1]

Solution

Let f(x) = log (x² + 2) - log 3
(a) Since log (x² + 2) and log 3 are both continuous in [-1, 1]
∴ 1og (x² + 2) - log 3 is continuous in [-1, 1] ⇒ f is continuous in [-1, 1]
$straight f apostrophe left parenthesis straight x right parenthesis equals fraction numerator 2 space straight x over denominator straight x squared plus 2 end fraction comma space which space exists space in space left parenthesis negative 1 comma space 1 right parenthesis$
∴ f is derivable in (-1, 1).
(c) f(-1) = log (1 + 2) - log 3 = 0
f(1) = log (1 + 2) - log 3 = 0
∴ f(-1) = f(1)
∴ f satisfies all the conditions of Rolle's Theorem
∴ there must exist at least one value c of x such that f'(c) = 0 where - 1 < c < 1
$Now space straight f apostrophe left parenthesis straight c right parenthesis equals 0 space gives space us space fraction numerator 2 space straight c over denominator straight c squared plus 2 end fraction equals 0 or space straight c equals 0 comma space element of left parenthesis negative 1 comma space 1 right parenthesis$
∴ Rolle's Theorem is verified.

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Continuity And Differentiability

Verify Roll�s Theorem for the function :f(x) = log (x² + 2) - log 3 in [-1, 1]

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For Daily Free Study Material Join wiredfaculty

Continuity And Differentiability

Verify Roll�s Theorem for the function :f(x) = log (x2 + 2) - log 3 in [-1, 1]

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Verify Roll�s Theorem for the function :f(x) = log (x² + 2) - log 3 in [-1, 1]