Answer:
Max = 86; min = 36.54
Step-by-step explanation:
[tex]f(x) = x^{2} + \dfrac{85}{x}[/tex]
Step 1. Find the critical points.
(a) Take the derivative of the function.
[tex]f'(x) = 2x - \dfrac{85}{x^{2}}[/tex]
Set it to zero and solve.
[tex]\begin{array}{rcl}2x - \dfrac{85}{x^{2}} & = & 0\\\\2x^{3} - 85 & = & 0\\2x^{3} & = & 85\\\\x^{3} & = &\dfrac{85}{2}\\\\x & = & \sqrt [3]{\dfrac{85}{2}}\\\\& \approx & 3.490\\\end{array}\[/tex]
(b) Calculate Æ’(x) at the critical point. Â
[tex]f(3.490) = 3.490^{2} + \dfrac{85}{3.490} = 12.18 + 24.36 = 36.54[/tex]
Step 2. Calculate Æ’(x) at the endpoints of the interval
[tex]f(1) = 1^{2} + \dfrac{85}{1} = 1 + 85 = 86\\\\f(5) = 5^{2} + \dfrac{85}{5} = 25 + 17 = 42[/tex]
Step 3.Identify the maxima and minima.
Æ’(x) achieves its absolute maximum of 86 at x = 1 and its absolute minimum of 36.54 at x = 3.490
The figure below shows the graph of Æ’(x) from x = 1 to x = 5.
