Determine the intervals on which the function is concave up or down and find the points of inflection.
$$f(x)=8x^3-5x^2+6$$
(Give your answer as a comma-separated list of points in the form $(\ast ,\ast$ ). Express numbers in exact form. Use notation and fractions where needed.)

Step 1 ：First, we find the first derivative of the function, which is $f^{\prime }(x)=24x^2-10x$.

Step 2 ：Next, we find the second derivative of the function, which is $f^{″}(x)=48x-10$.

Step 3 ：To find the points of inflection, we set the second derivative equal to zero and solve for x: $48x-10=0$, which simplifies to $x=5/24$.

Step 4 ：We then test the intervals around the point of inflection in the second derivative. For $x<5/24$, we use $x=0$ and find that $f^{″}(0)=-10$, which is less than 0. Therefore, the function is concave down on the interval $(-\infty ,5/24)$.

Step 5 ：For $x>5/24$, we use $x=1$ and find that $f^{″}(1)=38$, which is greater than 0. Therefore, the function is concave up on the interval $(5/24,\infty )$.

Step 6 ：Finally, we find that the point of inflection is at $(5/24,f(5/24))=(5/24,8\ast (5/24{)}^3-5\ast (5/24{)}^2+6)$.

Step 7 ：$\boxed{{\displaystyle \text{Therefore, the function is concave down on }(-\infty ,5/24)\text{ and concave up on }(5/24,\infty ).\text{ The point of inflection is at }(5/24,f(5/24))=(5/24,8\ast (5/24{)}^3-5\ast (5/24{)}^2+6)}}$