Is #f(x)=(-x^3-7x^2-x+2)/(x-2)# increasing or decreasing at #x=3#?

Question

Christopher Allers · Accepted Answer

#f(x)# increases at #x=3#

Determine whether the function's derivative is positive or negative at that point to ascertain whether a function is increasing or decreasing at that point.

First, differentiate #f(x)#

#f'(x) = d/dx (-x^3-7x^2-x+2)/(x-2)#

#= d/dx (-x^3-7x^2-x+2) * 1/(x-2)#

#= d/dx (-x^3-7x^2-x+2) * (x-2)^-1#

#= 1/(x-2) * d/dx (-x^3-7x^2-x+2) + (-x^3-7x^2-x+2) * d/dx (x-2)^-1#

#=1/(x-2) * (-3x^2-14x-1) + (-x^3-7x^2-x+2) * -(x-2)^-2 * d/dx (x-2)#

#=(-3x^2-14x-1)/(x-2) + (-x^3-7x^2-x+2) * -(x-2)^-2#

To put it simply,

#=(-3x^2-14x-1)/(x-2) - (-x^3-7x^2-x+2)/(x-2)^2#

#=((-3x^2-14x-1)(x-2))/(x-2)^2 - (-x^3-7x^2-x+2)/(x-2)^2#

#=(-3 x^3 - 8 x^2 + 27 x + 2 - (-x^3-7x^2-x+2))/(x-2)^2#

#=(-3 x^3 - 8 x^2 + 27 x + 2 +x^3 + 7x^2 + x - 2)/(x-2)^2#

#=(-2 x^3 - x^2 + 28 x)/(x-2)^2#

So to find out if the function is increasing or decreasing at #x=3#, find if #f'(3)# is positive or negative

#f'(3)=(-2 (3)^3 - (3) ^2 + 28 (3))/((3)-2)^2=21#

Since #f'(3)>0#, #f(x)# has a positive slope at x=3, and thus #f(x)# is increasing

graph{(x-2))=0 [1.423, 4.492, -91.816, -90.28]}

Samantha Adkison · Answer

undefined To determine if $f(x) = \frac{-x^3 - 7x^2 - x + 2}{x - 2}$ is increasing or decreasing at $x = 3$, we can use the first derivative test. Calculate $f'(x)$ and evaluate it at $x = 3$. If $f'(3) > 0$, then $f(x)$ is increasing at $x = 3$. If $f'(3) < 0$, then $f(x)$ is decreasing at $x = 3$.