Is #f(x)=(-2x^3-x^2-5x+2)/(x+1)# increasing or decreasing at #x=0#?
1 Answer
decreasing at x = 0
Explanation:
To test if a function is increasing / decreasing at x = a , require to check the sign of 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
Require to find f'(x)
differentiate 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 - x^2- 5x + 2 rArr g'(x) =-6x^2-2x-5# h(x) = x+1
# rArr h'(x) = 1#
#"------------------------------------------------------------------------"#
substitute these values into f'(x)
#rArr f'(x) =( (x+1).(-6x^2-2x-5) - (-2x^3-x^2-5x+2).1)/(x+1)^2# and f'(0) =
#(1.(-5) - 2.1)/1 = -7 # since f'(0) < 0 , f(x) is decreasing at x = 0
Here is the graph of f(x)
graph{(-2x^3-x^2-5x+2)/(x+1) [-10, 10, -5, 5]}