I will assume you meant
f(x) = x^3 + 3x^2 + 4
f ' (x) = 3x^2 + 6x = 0 for max/min
3x(x+2) = 0
x = 0 or x = -2
f(0) = 4
f(-2) = -8 + 12 + 4 = 8
end values:
f(-3) = -27 + 27 + 4 = 4
f(2) = 8 + 12 + 4 = 24
minimum value of the function is 4
maximum value is 24
Find the absolute maximum and absolute minimum values, if any, of the function. (If an answer does not exist, enter DNE.)
h(x) = x3 + 3x2 + 4 on [−3, 2]
1 answer
1 answer