ediate Value Theorem Quiz Let $f(x)$ be a function that is differentiable everywhere and has a derivative $f^{\prime}(x)=3 x^{2}+14 x+6$. Verify that the Intermediate Value Theorem for. Derivatives applies to the function $f^{\prime}(x)$ on the interval $[-6,-3]$, and find the value of $c$ guaranteed by the theorem such that $f^{\prime}(c)=-2$. (4 points)

Step 1 ：Calculate the derivative of the function at the endpoints of the interval [-6, -3]: \(f'(-6) = 3*(-6)^2 + 14*(-6) + 6 = 108 - 84 + 6 = 30\) and \(f'(-3) = 3*(-3)^2 + 14*(-3) + 6 = 27 - 42 + 6 = -9\)

Step 2 ：The derivative of the function takes on all values between -9 and 30 on the interval [-6, -3].

Step 3 ：Since -2 is between -9 and 30, by the Intermediate Value Theorem for Derivatives, there must be a value c in the interval [-6, -3] such that \(f'(c) = -2\).

Step 4 ：Set the derivative equal to -2 and solve for x: \(3x^2 + 14x + 6 = -2\) which simplifies to \(3x^2 + 14x + 8 = 0\)

Step 5 ：Solve the quadratic equation using the quadratic formula: \(x = [-14 ± \sqrt{(14)^2 - 4*3*8}] / (2*3)\), which simplifies to \(x = [-14 ± \sqrt{196 - 96}] / 6\), then to \(x = [-14 ± \sqrt{100}] / 6\), and finally to \(x = -4\) or \(x = -24/6 = -4\)

Step 6 ：The value of c that satisfies the conditions of the theorem is c = -4. However, -4 is not in the interval [-6, -3]. Therefore, there is no value of c in the interval [-6, -3] such that \(f'(c) = -2\).

Step 7 ：\(\boxed{\text{No solution}}\)