Question 1005100
f(x)=2/3 x^3-5x^2+12x-7
f'(x)=2x^2-10x+12
set it equal to zero and divide both sides by 2
x^2-5x+6=0
(x-3)(x-2)=0
x=3, 2  Critical points.
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f''(x)=4x-10
When x=3, f''(x) is positive, so a minimum
when x=2, f''(x) is negative, so a maximum
Set 4x-10=0, and x=2.5 is an inflection point.
Increasing at x>3, and concave downward from x>2.5
decreasing at x<2 and concave upward from x <2.5

{{{graph(300,200,-10,10,-10,10,(2/3)x^3-5x^2+12x-7)}}}