Let's consider the function $f(x)=2x^3-9x^2+12x-3$. Find the intervals of concavity and the inflection points.

Step 1 ：First, we find the first derivative of the function: $f^{\prime }(x)=6x^2-18x+12$

Step 2 ：Then, we find the second derivative of the function: $f^{″}(x)=12x-18$

Step 3 ：Set the second derivative equal to zero and solve for x: $12x-18=0$ which gives $x=1.5$

Step 4 ：To find the intervals of concavity, we test the sign of the second derivative on each interval. Choose test points $x=1$ and $x=2$. For $x=1$, $f^{″}(1)=-6<0$, so the function is concave down on the interval $(-\mathrm{\infty },1.5)$. For $x=2$, $f^{″}(2)=6>0$, so the function is concave up on the interval $(1.5,\mathrm{\infty })$

Step 5 ：The inflection point is then at $x=1.5$, and we find the corresponding y-value by substituting $x=1.5$ into the original function: $f(1.5)=2(1.5{)}^3-9(1.5{)}^2+12(1.5)-3=-1.125$. Therefore, the inflection point is $(1.5,-1.125)$