Is #f(x)=(-2x^3+4x^2-x-2)/(x+3)# increasing or decreasing at #x=-2#?
1 Answer
decreasing at x = -2
Explanation:
To determine if a function is increasing/decreasing at x = a we evaluate f'(a).
• If f'(a) > 0 , then f(x) is increasing at x = a
• If f'(a) < 0 , then f(x) is decreasing at x = a
differentiate f(x) using the
#color(blue)" quotient rule " # If
# f(x) = (g(x))/(h(x)) " then " f'(x) =(h(x).g'(x) - g(x).h'(x))/(h(x))^2 #
#"--------------------------------------------------------------"# g(x)
#-2x^3+4x^2-x-2 rArr g'(x) = -6x^2+8x-1 # h(x) = x + 3 → h'(x) = 1
#"----------------------------------------------------------"#
Substitute these values into f'(x)f'(x)
#=((x+3)(-6x^2+8x-1)-(-2x^3+4x^2-x-2).1)/(x+3)^2# and f'(-2)
#=(1.(-24-16-1)-(16+16+2-2))/1# = (-41-32) =-73
Since f'(-2) < 0 then f(x) is decreasing at x = -2
graph{(-2x^3+4x^2-x-2)/(x+3) [-10, 10, -5, 5]}
