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

c such that

f ′ (c ) = 0, if

f (x ) = x 2 - 4 x - 1 2 x + 4 in the interval

[- 2, 6]. B c A.

c - 4 - 2

5 B.

c 4

C.

c - 4 + 2

5 D.

c - 4

E.

c 0

c
C

Step 1: f (x ) = x 2 - 4 x - 1 2 x + 4 [Write the function.] Step 2: f (- 2) = ( - 2 ) 2 - 4 ( - 2 ) - 1 2 - 2 + 4 = 0[Find f (- 2).] Step 3: f (6) = ( 6 ) 2 - 4 ( 6 ) - 1 2 6 + 4 = 0[Find f (6).] Step 4: Since f (x ) is continuous in [- 2, 6] and differentiable in (- 2, 6) also f (- 2) = f (6) = 0, Rolle's Theorem is applicable. Step 5: There exists a number c in (- 2, 6) such that f ′(c ) = 0. Step 6: f ′(c ) = c 2 + 8 c - 4 ( c + 4 ) 2 = 0
[Find f ′(c ).] Step 7: c 2 + 8 c - 4 ( c + 4 ) 2 = 0[Equate f ′(c ) to zero.] Step 8: c ^{2} + 8c - 4 = 0 Step 9: c = - 4 + 25 , which lies between the interval (- 2, 6)[Solve for c .] Step 10: So, the derivative of the function is zero at c = - 4 + 25 in the interval [- 2, 6].