Is #f(x)=(4x^3-2x^2-x-3)/(x-2)# increasing or decreasing at #x=0#?
1 Answer
increasing at x = 0
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)
#=4x^3-2x^2-x-3 rArr g'(x) = 12x^2 -4x - 1 # and h(x) = x-2
#rArr h'(x) = 1 #
#"---------------------------------------------------------------------"#
substitute these values into f'(x)f'(x)
#= ((x-2)(12x^2-4x-1) - (4x^3-2x^2-x-3).1)/(x-2)^2 # and f'(0)
# = ((-2)(-1) - (-3))/(-2)^2 = 5/4 # Since f'(0) > 0 , then f(x) is increasing at x = 0
graph{(4x^3-2x^2-x-3)/(x-2) [-10, 10, -5, 5]}
