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 Attempt following question by selecting a choice to answer. eUse Rolle's Theorem to determine the value of c such that f ′ (c) = 0, if f(x) = x(x + 3) e-x2 in the interval [- 3, 0].ee   eeA.  e- 2B.  e0C.  e- 2.5D.  e- 1E.  e- 3e A Step 1: f(x) = x(x + 3) e-x2[Write the function.]Step 2: f(- 3) = (- 3)(- 3 + 3) e32 = 0[Find f(- 3).]Step 3: f(0) = 0[Find f(0).]Step 4: Since f(x) is continuous in [- 3, 0] and differentiable in (- 3, 0) also f(- 3) = f(0) = 0, Rolle's Theorem is applicable. Step 5: There exists a number c in (- 3, 0) such that f ′(c) = 0Step 6: f(x) = x(x + 3) e-x2 ⇒ f ′(x) = 12 e-x2 (- x2 + x + 6)[Find f ′ (x).]Step 7: f ′(c) = 12 e-c2 (- c2 + c + 6)[Find f ′(c).] Step 8: 12 e-c2 (- c2 + c + 6) = 0 [Equate f ′(c) to zero.]Step 9: - c2 + c + 6 = 0 [e-c2 > 0 for all c.]Step 10: c = - 2, 3[Solve for c.]Step 11: c = - 2, which lies in the interval (- 3, 0).Step 12: So, the derivative of the function is zero at c = - 2 in the interval [- 3, 0].