The function ƒ(x) = x3 – 3x + 2 is increasing on the interval -1 < x < 1.
True or False
x3 means x^3
f'(x) = 3x^2-3 = 3(x^2-1)
f increases if f' > 0
so, what do you think?
To determine whether the function ƒ(x) = x^3 - 3x + 2 is increasing on the interval -1 < x < 1, we need to find the derivative of the function and check its sign within that interval.
First, let's find the derivative of ƒ(x) with respect to x:
ƒ'(x) = d/dx (x^3 - 3x + 2)
To find the derivative, we can apply the power rule and the constant rule of differentiation:
ƒ'(x) = 3x^2 - 3
Now, we need to determine the sign of the derivative within the interval -1 < x < 1. To do this, we can evaluate the derivative at any value within that interval. Let's choose x = 0:
ƒ'(0) = 3(0)^2 - 3
ƒ'(0) = -3
Since ƒ'(0) is negative, this means that the derivative is negative within the interval -1 < x < 1. When the derivative is negative, it indicates that the function is decreasing.
Therefore, the statement "The function ƒ(x) = x^3 - 3x + 2 is increasing on the interval -1 < x < 1" is False.