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

c such that

f ′ (c ) = 0, if

f (x ) = (x + 3)(x - 1)^{2} in the interval

[- 3, 1]. D B A.

B - 1

B.

B 1

C.

B 5 3 D.

B 0

E.

B -

5 3 B
E

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