Is #f(x)=(-3x^2-x+2)/(x^2+x)# increasing or decreasing at #x=1#?
1 Answer
decreasing at x = 1
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)
#= -3x^2-x+2 rArr g'(x) = -6x - 1 # and h(x)
#=x^2 + x rArr h'(x) = 2x + 1 #
#"---------------------------------------------------------------"#
Substitute these values into f'(x)
# f'(x) = ((x^2+x)(-6x-1) - (-3x^2-x+2)(2x+1))/(x^2+x)^2 # and f'(1)
#= (2.(-7) - (-2).(3))/2^2 = (-8)/4 = -2# Since f'(1) < 0 , then f(x) is decreasing at x = 1
graph{(-3x^2-x+2)/(x^2+x) [-10, 10, -5, 5]}
