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 Attempt following question by selecting a choice to answer. BUse Rolle's Theorem to determine the value of c such that f ′(c) = 0, if f(x) = log (x2+12-2x) in the interval [- 4, - 3].eee  BBBA.  B14B.  B- 23C.  B- 22D.  B- 43E.  B- 14BB B Step 1: f(x) = log (x2+12-2x)[Write the function.]Step 2: f(- 4) = log ((-4)2+12-2(-4)) = 0.544[Find f(- 4).]Step 3: f(- 3) = log ((-3)2+12-2(-3)) = 0.544[Find f(- 3).]Step 4: Since f(x) is continuous in [- 4, - 3] and differentiable in (- 4, - 3) also f(- 4) = f(- 3) = 0.544, Rolle's Theorem is applicable.Step 5: There exists a number c in (- 4, - 3) such that f ′(c) = 0 Step 6: f(x) = log (x2+12-2x) ⇒f ′(x) = x2-12x(x2+12)[Find f ′(x).]Step 7: f ′(c) = c2-12c(c2+12)[Find f ′(c).]Step 8: c2-12c(c2+12) = 0 [Equate f ′(c) to zero.]Step 9: c2 - 12 = 0Step 10: c = - 23, 23[Solve for c.]Step 11: c = - 23, which lies in the interval (- 4, - 3)Step 12: So, the derivative of the function is zero at c = - 23 in the interval [- 4, - 3].