calc
posted by casey .
let f(x)= x^2 +3x on the interval [1,3] . Find the absolute maximum and absolute minimum of f(x) on this interval

The absolute maximum and absolute minimum of implicit equations are difficult to find because of the uncertainty of the number of local maxima and minima.
For a quadratic equation, we know that there is only one single maximum/minimum which will automatically be the global value. If a global maximum falls within the interval, the smaller value of each of the limits will give the minimum within the interval.
Thus, interval = [1,3].
f(x)= x^2 +3x
f'(x)=2x+3
f'(x)=0 at x=1.5, falls within [1,3]
f"(x)=2 f'(1.5) is a maximum.
Thus the maximum is at x=1.5, or f(1.5)=2.25.
The minimum is one of the two following values (evaluated at the limits of the given interval).
f(1) or f(3).
Can you take it from here?