Answer:
x=1
Step-by-step explanation:
We are given that
[tex]\frac{2x^3}{3}+2x^2-6x+4[/tex]
We have to find the value of x for which function is minimum.
Differentiate w.r.t x
[tex]f'(x)=2x^2+4x-6[/tex]
Substitute f'(x)=0
[tex]2x^2+4x-6=0[/tex]
[tex]x^2+2x-3=0[/tex]
[tex]x^2+3x-x-3=0[/tex]
[tex]x(x+3)-1(x+3)=0[/tex]
[tex](x-1)(x+3)=0[/tex]
[tex]x=1,-3[/tex]
Again, differentiate w.r.t x
[tex]f''(x)=4x+4[/tex]
[tex]f''(1)=4(1)+4=8>0[/tex]
[tex]f''(-3)=4(-3)+4=-8<0[/tex]
Hence, the function is minimum at x=1
