Identify where the function $f(x)=x^{3}-5 x^{2}+10 x+1$ is increasing, decreasing, concave up, and concave down.
(Give your answer as an interval in the form $(*, *)$. Use the symbol $\infty$ for infinity, $\cup$ for combining intervals, and an appropriate type of parentheses "(",")", "[", or "]" depending on whether the interval is open or closed. Express numbers in exact form. Use symbolic notation and fractions where needed. Enter DNE if no such interval exists.)
$f$ is increasing on:
$f$ is decreasing on:
$f$ is concave up on:

Step 1 ：Find the derivative of the function, \(f'(x) = 3x^{2}-10x+10\)

Step 2 ：Set the derivative equal to zero to find critical points, \(3x^{2}-10x+10 = 0\)

Step 3 ：Solve the quadratic equation using the quadratic formula, \(x = \frac{10 \pm \sqrt{(-10)^{2}-4*3*10}}{2*3} = \frac{10 \pm \sqrt{100-120}}{6} = \frac{10 \pm \sqrt{-20}}{6}\)

Step 4 ：Since the square root of a negative number is not a real number, there are no real solutions to this equation. Therefore, the function does not have any critical points, and it is either always increasing or always decreasing.

Step 5 ：Take the second derivative of the function, \(f''(x) = 6x-10\)

Step 6 ：The function is increasing when \(f''(x) > 0\), which is when \(x > \frac{10}{6} = \frac{5}{3}\)

Step 7 ：The function is decreasing when \(f''(x) < 0\), which is when \(x < \frac{5}{3}\)

Step 8 ：Set the second derivative equal to zero to find inflection points, \(6x-10 = 0\)

Step 9 ：Solve for \(x\), we get \(x = \frac{10}{6} = \frac{5}{3}\)

Step 10 ：The function is concave up on the interval \((-\infty, \frac{5}{3})\) and concave down on the interval \((\frac{5}{3}, \infty)\)

Step 11 ：\(\boxed{f \text{ is increasing on: } (\frac{5}{3}, \infty)}\)

Step 12 ：\(\boxed{f \text{ is decreasing on: } (-\infty, \frac{5}{3})}\)

Step 13 ：\(\boxed{f \text{ is concave up on: } (-\infty, \frac{5}{3})}\)

Step 14 ：\(\boxed{f \text{ is concave down on: } (\frac{5}{3}, \infty)}\)