I) Find the first derivative -> f'(x) = -3x^2 + 14x - 15 and set it equal to zero to solve. This will find the x-values of your "critical points". A critical point is where the graph humps. Solving, you get x = 5/3 and x = 3, meaning there is a hump (relative extrema) at those values. Because we don't/can't graph the function, we don't know if those humps are hills or valleys. If necessary, you can find the exact point of those critical points by plugging 5/3 into the original function to determine the y-value (and likewise for x = 3).
II) Find the second derivative -> f''(x) = -6x + 14. The second derivative will help you find the exact point on a graph where a graph changes from a hill to a valley (or vise versa). This concept is called "concavity." Plug the critical points from Part One into the second derivative: f''(5/3) = a positive number -- the number four to be exact, but that won't matter to you -- and f''(3) = a negative number -- again, the number negative four to be exact, but that won't matter to you. Why does the exact value not matter to you, you ask? If the second derivative gives you a positive number, then your graph is "concave up" (or CUP UP or a VALLEY) and if the second derivative gives you a negative number, then your graph is "concave down" (or CUP DOWN or a HILL). [A cup that is up would make a valley, but a cup that is upside down would make a hill].
Find the first derivative.
Now we need to find the value(s) of x that make the first derivative zero to find the critical points.
`x=5/3 ` and `x=3 ` (critical points)
These will be a relative extrema if it changes sign, so find 2 values around both to test using the first derivative test, like 1 and 2 for the first and 2 and 3 for the second.
Therefore, this is a relative extrema. Find the y-value, which is -10.19. `(5/3,-10.19) `
Do the same for the other critical point
This is also a relative extrema. Find the y-value, which is -9. `(3,-9) `