f(-3) = -1
f(1) = 3
Now we have the max/min at the endpoints. There may be other, different min/max inside the interval.
f'(x) = 3x^2 + 6x = 3x(x+2)
so,
f(-2) = 3 is a local max
f(0) = -1 is a local min
On the interval [-3,1] the absolute
max is 3
min is -1
Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)
f(x) = x^3 + 3 x^2 - 1 on [-3, 1]
1 answer
1 answer