On what interval is the function f(x)=x^3-4x^2+5x concave upward?

Calculus 1?

To find concavity you must take the second derivative.  

As you would to find your local maximums and minimums (critical points) in the first derivative by setting y' = 0, to find points of inflection you set acceleration, y" = 0.  

Now that you know where the point in which the function is neither concave up or concave down (at the points of inflection) plug x-values between them into the second derivative for x.  If y" is positive between those particular points will be concave up and if y" is negative it will be concave down between that interval.

For a better understanding you might find a good video on Youtube explaining this if you search "Points of Inflections" or "Concavity of a function".

Cheers.

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