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Chapter 7: Problem 3

The set of all values of the parameters a for which the points of minimum of the function \(y=1+a^{2} x-x^{3}\) satisfy the inequality \(\frac{x^{2}+x+2}{x^{2}+5 x+6} \leq 0\) is (A) an empty set (B) \((-3 \sqrt{3},-2 \sqrt{3})\) (C) \((2 \sqrt{3}, 3 \sqrt{3})\) (D) \((-3 \sqrt{3},-2 \sqrt{3}) \cup(2 \sqrt{3}, 3 \sqrt{3})\)

Short Answer

Expert verified
Answer: The set of values of the parameter a is an empty set.

Step by step solution

01

Find the critical points of the function

To find the points of minima, first find the critical points of the function y. Do this by taking the first derivative of y with respect to x, and setting it equal to 0. $$\frac{dy}{dx} = 2 a^{2} x - 3 x^{2}$$ Then, set \(\frac{dy}{dx}=0\) and solve for x: $$2 a^{2} x - 3 x^{2} = 0$$ $$x ( 2a^{2} - 3x) = 0$$ The critical points are thus x = 0, and \(x = \frac{2a^2}{3}\).
02

Determine which critical point gives a minimum

To determine which critical point gives a minimum, consider the second derivative test. Calculate the second derivative of y with respect to x: $$\frac{d^{2}y}{dx^{2}} = 2 a^{2} - 6 x$$ Now, plug in the critical points x = 0 and \(x = \frac{2a^2}{3}\): $$\frac{d^{2}y}{dx^{2}} (0) = 2 a^{2}$$ $$\frac{d^{2}y}{dx^{2}} \left(\frac{2a^2}{3}\right) = 2 a^{2} - 4a^2 = -2 a^{2}$$ Since \(\frac{d^{2}y}{dx^{2}} (0) > 0\) (minimum) and \(\frac{d^{2}y}{dx^{2}} \left(\frac{2a^2}{3}\right) < 0\) (maximum), the only critical point corresponding to a minimum is x = 0.
03

Determine the set of parameter a values from the inequality

Now, plug the minimum value of x (which is x=0) into the inequality and solve for a: $$\frac{0^{2}+0+2}{0^{2}+5\cdot0+6} \leq 0$$ This inequality is not true because the left side equals \(\frac{2}{6}\), which is greater than 0. Therefore, the set of parameter a values for which the points of the minimum of the function satisfy the inequality is the empty set. Hence, the answer is: (A) an empty set

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