We develop the expression and we calculate the first and second derivatives
f(x)=(2x-1)(3x-5)(x-2)
=(6x^2-13x+5)(x-2)
=6x^3-12x^2-13x^2+26x+5x-10
f(x)=6x^3-25x^2+31x-10
f'(x)=18x^2-50x+31
f''(x)=36x-50
f''(x)=0 when x=50/36=25/18
We construct a table
color(white)(aaaa)Intervalcolor(white)(aaaa)|color(white)(aaaa)]-oo,25/18[color(white)(aaaa)|color(white)(aaa)]25/18,+oo[|
color(white)(aaaa)Sign f''(x)color(white)(aa)|color(white)(aaaaaaaa)-color(white)(aaaaaaa)|color(white)(aaaaaaaa)+color(white)(aa)|
color(white)(aaaa)functioncolor(white)(aaaa)|color(white)(aaaaaaaaa)nnncolor(white)(aaaaaa)|color(white)(aaaaaaaa)uuucolor(white)(aa)|
Therefore,
f(x) is concave when x in ]-oo,25/18[ and convex when x in ]25/18,+oo[