Is #f(x)=(x-2)(x+5)(x+2)# increasing or decreasing at #x=-3#?

Question

Charles Anctil · Accepted Answer

#f(x)=(x-2)(x+5)(x+2)# is decreasing at #x=-3# We are given #f(x)=(x-2)(x+5)(x+2)# Now the function is increasing at a given point, if #(df)/(dx)>0# at that point and is decreasing if #(df)/(dx)<0#. Here #f(x)=(x-2)(x+5)(x+2)=(x+5)(x^2-4)=x^3+5x^2-4x-20# and #(df)/(dx)=3x^2+10x-4# at #x=-3#, we have #((df)/(dx))_(x=-3)=3(-3)^2+10(-3)-4# = #27-30-4=-7# Hence #f(x)=(x-2)(x+5)(x+2)# is decreasing at #x=-3# (see graph not drawn to scale - shrunk alon #y#-axis) graph{(x-2)(x+5)(x+2) [-10, 10, -40, 40]}

Charles Anctil · Answer

undefined To determine if the function $ f(x) = (x - 2)(x + 5)(x + 2) $ is increasing or decreasing at $ x = -3 $, we need to analyze the sign of the derivative at that point.

1. Compute the derivative of $ f(x) $ using the product rule.
2. Evaluate the derivative at $ x = -3 $.
3. If the derivative is positive at $ x = -3 $, the function is increasing at that point. If it's negative, the function is decreasing.

Let's perform these steps:

1. $ f'(x) = (x + 5)(x + 2) + (x - 2)(x + 2) + (x - 2)(x + 5) $.
2. $ f'(-3) = (2)(-1) + (-5)(-1) + (-5)(2) = -2 + 5 - 10 = -7 $.
3. Since $ f'(-3) = -7 $ (negative), the function is decreasing at $ x = -3 $.