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Attempt following question by selecting a choice to answer. D Use Rolle's Theorem to determine the value of

c such that

f ′(c ) = 0, if

f (x ) = log (x 2 + 1 2 - 2 x ) in the interval

[- 4, - 3]. BBBB DDDD A.

D 1 4 B.

D - 2

3 C.

D - 2

2 D.

D - 4

3 E.

D -

1 4 DD
B

Step 1: f (x ) = log (x 2 + 1 2 - 2 x )[Write the function.] Step 2: f (- 4) = log (( - 4 ) 2 + 1 2 - 2 ( - 4 ) ) = 0.544[Find f (- 4).] Step 3: f (- 3) = log (( - 3 ) 2 + 1 2 - 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 (x 2 + 1 2 - 2 x ) ⇒ f ′(x ) = x 2 - 1 2 x ( x 2 + 1 2 ) [Find f ′(x ).] Step 7: f ′(c ) = c 2 - 1 2 c ( c 2 + 1 2 ) [Find f ′(c ).] Step 8: c 2 - 1 2 c ( c 2 + 1 2 ) = 0
[Equate f ′(c ) to zero.] Step 9: c ^{2} - 12 = 0 Step 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].