[tex]f(x) = x^3+x^2-x[/tex]
Finding the extreme point :
[tex]f'(x)=3x^2+2x-1=0\\\\3x^2+3x-x-1=0\\\\3x(x+1)-1(x+1)=0\\\\(3x-1)(x+1)=0\\\\x=\dfrac{1}{3}\ ,\ x=-1[/tex]
Now,
f''( x ) at x = 1/3 and -1 is :
[tex]f''(x)=6x+2\\\\f''(1/3)=4 \\\\f''(-1)=-4[/tex]
Therefore, it is maximum at x = -1 and minimum at x = 1/3 .
Hence, this is the required solution.
