Problem 61 - Essential Calculus

Question

Find the maximum value of \(f(x)=x^{3}-3 x\) on the set of all real numbers \(x\) satisfying \(x^{4}+36 \leq 13 x^{2}\). Explain your reasoning.

Step-by-Step Solution

Answer

The maximum value of the function on the given domain is 8.

Step 1: Solve the inequality

To start off with, rewrite the inequality \(x^{4}+36 \leq 13 x^{2}\) in standard form: \(x^{4}-13x^{2}+36 \leq 0\). Factorize to obtain \( (x^{2}-4)(x^{2}-9) \leq 0 \). Consequently, the roots of this equation are \(x=2, -2, 3, -3\). Now, the solution set for the inequality can be obtained by testing the intervals \((- \infty, -3), (-3, -2), (-2, 2), (2,3), (3, \infty)\) and determining for which intervals the function \( (x^{2}-4)(x^{2}-9)\) is negative or zero. This gives the solution set \(-3 \leq x \leq -2, 2 \leq x \leq 3\)

Step 2: Find the critical points

To find the critical points of \(f(x)=x^{3}-3x\), set its derivative equal to zero and solve for \(x\). The derivative is \(f'(x)=3x^{2}-3\). Set this to zero and solve for \(x\), gives \(x=\sqrt{1}, -\sqrt{1}\) which translates into the values \(x = -1,1\). However, among these solutions, only \(x=1\) falls in the determined domains from step 1.

Step 3: Evaluate the function at the endpoints and at the critical point

Next, evaluate \(f(x)=x^{3}-3x\) at the endpoints \(-3, -2, 2, 3\) and the critical point \(1\). The maximum value will be the maximum of these 5 numbers. The function yields the following values at the endpoints and the critical points: \(f(-3) = 0, f(-2) = -8, f(2) = 8, f(3) = 0, f(1) = -2\). Therefore, the maximum value is \(8\).

Step 4: Conclusion

The maximum value of the function \(f(x)=x^{3}-3x\) on the set of all real numbers \(x\) satisfying \(x^{4}+36 \leq 13x^{2}\) is \(8\) and occurs when \(x = 2\).