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Question: Given x>0 x > 0 , use the basic inequality to find \[ \left( \frac{3x - 1}{x(x + 1)}…

Given x>0 x > 0 , use the basic inequality to find (3x1x(x+1))max \left( \frac{3x - 1}{x(x + 1)} \right)_{\text{max}}

Solution

To find the maximum value of 3x1x(x+1) \frac{3x - 1}{x(x+1)} given x>0x > 0, we’ll use calculus by taking the derivative and finding critical points. First, let f(x)=3x1x(x+1) f(x) = \frac{3x - 1}{x(x+1)} Rewrite the function: f(x)=3x1x2+x f(x) = \frac{3x - 1}{x^2 + x} Differentiate f(x)f(x) using the quotient rule: f(x)=(x2+x)(3x1)(3x1)(x2+x)(x2+x)2 f’(x) = \frac{(x^2+x)‘(3x-1) - (3x-1)‘(x^2+x)}{(x^2+x)^2} Find the derivatives: (x2+x)=2x+1 (x^2 + x)’ = 2x + 1 (3x1)=3 (3x - 1)’ = 3 Substitute back: f(x)=(2x+1)(3x1)3(x2+x)(x2+x)2 f’(x) = \frac{(2x+1)(3x-1) - 3(x^2+x)}{(x^2+x)^2} Simplify the numerator: =(2x+1)(3x1)3x23x = (2x+1)(3x-1) - 3x^2 - 3x =(6x22x+3x1)3x23x = (6x^2 - 2x + 3x - 1) - 3x^2 - 3x =6x2+x13x23x = 6x^2 + x - 1 - 3x^2 - 3x =3x22x1 = 3x^2 - 2x - 1 Set the derivative to zero to find critical points: 3x22x1=0 3x^2 - 2x - 1 = 0 Use the quadratic formula: x=b±b24ac2a x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} where a=3a = 3, b=2b = -2, c=1c = -1. x=(2)±(2)24×3×(1)2×3 x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \times 3 \times (-1)}}{2 \times 3} x=2±4+126 x = \frac{2 \pm \sqrt{4 + 12}}{6} x=2±166 x = \frac{2 \pm \sqrt{16}}{6} x=2±46 x = \frac{2 \pm 4}{6} Possible solutions are: x=1andx=13 x = 1 \quad \text{and} \quad x = -\frac{1}{3} Since x>0x > 0, consider x=1x = 1. Evaluate ff at the endpoint and critical point: f(1)=3(1)11(1+1)=22=1 f(1) = \frac{3(1) - 1}{1(1+1)} = \frac{2}{2} = 1 Therefore, the maximum value of 3x1x(x+1) \frac{3x - 1}{x(x+1)} for x>0x > 0 is 1.

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