Is #f(x)=(x-1)(x-3)(2x-1)# increasing or decreasing at #x=2#?

1 Answer
Oct 1, 2016

It's decreasing.

Explanation:

So, to determine whether a function is increasing or decreasing at a point algebraically, the way to go is with a derivative. So, we want to find #f'(x)#, or the derivative of #f(x)#. We could use a product rule, but frankly it's much easier to just foil and use power rule.

I'll save you the work and tell you the the foiled version of #f(x)# is #2x^3 - 9x^2 + 10x - 3#

Now, this becomes a very simple power rule derivative. It will end up as:
#f'(x) = 6x^2 - 18x + 10#

Before we go any further, we need to establish what the derivative will tell us. Now, by definition, the derivative is the instantaneous rate of change, or the slope of the tangent line at a point. How does this help us? Well, if our derivative is positive at a point, then what would that imply? Well, it would imply that the instantaneous rate of change at that point is positive, or that the function is increasing there. We can use the same logic to say that if the derivative is negative at a point, the function is decreasing there.

Now, all that is left to do is actually take the derivative at #x=2#
Let's do that:

#f'(2) = 6(2)^2 -18(2) + 10#
#= 24 - 36 +10#
#= -2#

So, as #f'(x) < 0# (i.e. the derivative is negative) at #x=2#, we can say that #f(x)# is decreasing at #x=2#.

Here's a graph of the function as well, that shows us that we are correct:

graph{(x-1)(x-3)(2x-1) [-10, 10, -5, 5]}

Hope that helped :)