Is #f(x)=(x^3-4x^2+3x-4)/(4x-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 ,consider the following.
• 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"#
#f(x)=(g(x))/(h(x))" then "f'(x)=(h(x).g'(x)-g(x).h'(x))/(h(x))^2#
#"--------------------------------------------------------------------"#
#g(x)=x^3-4x^2+3x-4rArrg'(x)=3x^2-8x+3#
#h(x)=4x-2rArrh'(x)=4#
#"-------------------------------------------------------------------"#
Substitute these values into f'(x)
#f'(x)=((4x-2)(3x^2-8x+3)-(x^3-4x^2+3x-4).4)/(4x-2)^2# and
#f'(0)=((-2)(+3)-(-4).4)/(-2)^2=5/2# Since f'(0) > 0 , then f(x) is increasing at x = 0
graph{(x^3-4x^2+3x-4)/(4x-2) [-10, 10, -5, 5]}