How do you find the exact relative maximum and minimum of the polynomial function of #f(x) = x^3 + x + 4/x#?

Question

Luke Alvarez · Accepted Answer

# f_min=f(1)=6, and, f_max=f(-1)=-6#.

We know that, for # (1) :f_max, f'(x)=0 and f''(x) lt 0, and, # # (2) : f_min, f'(x)=0, and, f''(x) gt 0#. Now, #f(x)=x^3+x+4/x#, # rArr f'(x)=3x^2+1-4/x^2, and, f''(x)=6x+8/x^3#. #:. f'(x)=0 rArr 3x^2+1-4/x^2=0, or, 3x^4+x^2-4=0#. #:. (x^2-1)(3x^2+4)=0#. #because AA x in RR, 3x^2+4 !=0, x^2-1=0, or, x=+-1#. Then, #f''(1)=6+8=14 gt 0, f_min=f(1)=6#. Further, #f''(-1)=-14 lt 0, f_max=f(-1)=-6#. Spread the Joy of Maths.!

Charles Arocha · Answer

undefined To find the exact relative maximum and minimum of the polynomial function $ f(x) = x^3 + x + \frac{4}{x} $, follow these steps: 1. Find the derivative of the function $ f'(x) $. 2. Set $ f'(x) $ equal to zero and solve for $ x $ to find critical points. 3. Determine the second derivative $ f''(x) $. 4. Evaluate $ f''(x) $ at each critical point. 5. If $ f''(x) > 0 $, the function has a relative minimum at that point. 6. If $ f''(x) < 0 $, the function has a relative maximum at that point. Let's solve it step by step: 1. $ f(x) = x^3 + x + \frac{4}{x} $ $ f'(x) = 3x^2 + 1 - \frac{4}{x^2} $ 2. Set $ f'(x) = 0 $: $ 3x^2 + 1 - \frac{4}{x^2} = 0 $ $ 3x^4 + x^2 - 4 = 0 $ $ (x^2 + 4)(3x^2 - 1) = 0 $ $ x^2 = -4 $ or $ x^2 = \frac{1}{3} $ Since $ x^2 = -4 $ has no real solutions, we consider $ x^2 = \frac{1}{3} $. $ x = \pm \frac{1}{\sqrt{3}} $ 3. Find $ f''(x) $: $ f''(x) = 6x + \frac{8}{x^3} $ 4. Evaluate $ f''(x) $ at critical points: $ f''\left(\frac{1}{\sqrt{3}}\right) = 6\left(\frac{1}{\sqrt{3}}\right) + \frac{8}{\left(\frac{1}{\sqrt{3}}\right)^3} $ $ f''\left(\frac{1}{\sqrt{3}}\right) > 0 $ (positive), so relative minimum exists at $ x = \frac{1}{\sqrt{3}} $. $ f''\left(-\frac{1}{\sqrt{3}}\right) = 6\left(-\frac{1}{\sqrt{3}}\right) + \frac{8}{\left(-\frac{1}{\sqrt{3}}\right)^3} $ $ f''\left(-\frac{1}{\sqrt{3}}\right) < 0 $ (negative), so relative maximum exists at $ x = -\frac{1}{\sqrt{3}} $. Therefore, the exact relative maximum occurs at $ x = -\frac{1}{\sqrt{3}} $, and the exact relative minimum occurs at $ x = \frac{1}{\sqrt{3}} $.