Question: Given x>0, use the basic inequality to find \[ \left( \frac{3x - 1}{x(x + 1)}…
Given x>0, use the basic inequality to find
(x(x+1)3x−1)max
Solution
To find the maximum value of x(x+1)3x−1 given x>0, we’ll use calculus by taking the derivative and finding critical points. First, let f(x)=x(x+1)3x−1 Rewrite the function: f(x)=x2+x3x−1 Differentiate f(x) using the quotient rule: f’(x)=(x2+x)2(x2+x)‘(3x−1)−(3x−1)‘(x2+x) Find the derivatives: (x2+x)’=2x+1(3x−1)’=3 Substitute back: f’(x)=(x2+x)2(2x+1)(3x−1)−3(x2+x) Simplify the numerator: =(2x+1)(3x−1)−3x2−3x=(6x2−2x+3x−1)−3x2−3x=6x2+x−1−3x2−3x=3x2−2x−1 Set the derivative to zero to find critical points: 3x2−2x−1=0 Use the quadratic formula: x=2a−b±b2−4ac where a=3, b=−2, c=−1. x=2×3−(−2)±(−2)2−4×3×(−1)x=62±4+12x=62±16x=62±4 Possible solutions are: x=1andx=−31 Since x>0, consider x=1. Evaluate f at the endpoint and critical point: f(1)=1(1+1)3(1)−1=22=1 Therefore, the maximum value of x(x+1)3x−1 for x>0 is 1.