To find the local maximum of the function [tex]g(x)=x^3+5x^2-17x-21,[/tex] you should:
1. find the derivative [tex]g'(x)=3x^2+10x-17;[/tex]
2. find stationary points. Equate derivative to zero and then solve the equation
[tex]3x^2+10x-17=0,\\ \\D=10^2-4\cdot 3\cdot (-17)=100+204=304,\\ \\\sqrt{D}=4\sqrt{19} ,\\ \\x_{1,2}=\dfrac{-10\pm 4\sqrt{19}}{2\cdot 3}=\dfrac{-5\pm 2\sqrt{19}}{3}.[/tex]
3. Determine signs of g'(x):
- when [tex]x<\dfrac{-5-2\sqrt{19}}{3},[/tex] then g'(x)>0 (function g(x) is increasing);
- when [tex]\dfrac{-5-2\sqrt{19}}{3}<x<\dfrac{-5+2\sqrt{19}}{3},[/tex] then g'(x)<0 (function g(x) is decreasing);
- when [tex]x>\dfrac{-5+2\sqrt{19}}{3},[/tex] then g'(x)>0 (function g(x) is increasing).
4. This means that [tex]x=\dfrac{-5-2\sqrt{19}}{3}[/tex] is point of maximum and [tex]x=\dfrac{-5+2\sqrt{19}}{3}[/tex] is point of minimum.
5. The maximum value of g(x) is at [tex]x=\dfrac{-5-2\sqrt{19}}{3}:[/tex]
[tex]g\left(\dfrac{-5-2\sqrt{19}}{3}\right)=\left(\dfrac{-5-2\sqrt{19}}{3}\right)^3+5\left(\dfrac{-5-2\sqrt{19}}{3}\right)^2-17\left(\dfrac{-5-2\sqrt{19}}{3}\right)-21\approx 65.6705658\approx 65.671[/tex]
