Is #f(x)=-2x^3+2x^2-x+2# increasing or decreasing at #x=0#?

Question

Charles Arocha · Accepted Answer

Hey there! Simply put, #f(x)# is decreasing at x = 0. I'll explain why below!

Finding a characteristic of this function at x = 0—that is, if it is increasing at a certain point—will require you to perform what is referred to as the First Derivative Test. Like with any function, you need to know the first derivative to figure out if the function is increasing or decreasing at a given point. When we differentiate, we obtain: #f'(x) = -6x^2 +4x - 1 # The first derivative needs to be substituted with the value of x that you currently have. # f'(0) = -6(0)^2 +4(0) - 1 # Simplifying, we obtain: #f'(0) = -1# What does the first derivative mean, and how do we interpret this? A function's slope or rate of change is represented by its first derivative, and if the slope is negative, the function is decreasing; on the other hand, if the first derivative is positive, the function is increasing! I hope this was helpful. Please feel free to ask any questions you may have, and I will try my best to respond. :)

Cameron Alexander · Answer

undefined To determine whether the function $ f(x) = -2x^3 + 2x^2 - x + 2 $ is increasing or decreasing at $ x = 0 $, we need to analyze the sign of the derivative of the function at that point.

First, find the derivative of $ f(x) $ with respect to $ x $:

$$ f'(x) = -6x^2 + 4x - 1 $$

Now, evaluate $ f'(0) $:

$$ f'(0) = -6(0)^2 + 4(0) - 1 = -1 $$

Since $ f'(0) $ is negative, the function is decreasing at $ x = 0 $.