you know that any local min/max must occur where f' = 0.
f' = ln5 * 5^x - 4x^2
That occurs at about x=-.5 and 1.24
So, check f(x) at the endpoints of the interval, and at those two values.
Looks to me like max is at x=1.24, min at x=2
Treat the other similarly.
1.Determine the maximum value of the function f(x)= 5^x -x^5 on the interval 0≤x≤≤2.
2. Determine the minimum value of the function f(x)= e^(-x) - 2e^x on the interval -1≤x≤2.
1 answer
1 answer