Is #f(x)=(-x^3+x^2-5x+6)/(x-2)# increasing or decreasing at #x=0#?
1 Answer
Jan 26, 2016
increasing at x = 0
Explanation:
To test if f(x) is increasing / decreasing :
• If f'(x) > 0 at x = 0 then increasing
• If f'(x) < 0 at x = 0 then decreasingDifferentiate f(x) using
#color(blue)(" quotient rule") #
# f'(x) =((x-2)d/dx(-x^3+x^2-5x+6)-(-x^3+x^2-5x+6)d/dx(x-2))/(x-2)^2#
# =( (x-2)(-3x^2+2x-5) -(-x^3+x^2-5x+6) .1)/(x-2)^2#
# =( -3x^3+2x^2-5x+6x^2-4x+10+x^3-x^2+5x-6)/(x-2)^2#
# =( -2x^3+7x^2 -4x+4)/(x-2)^2# # f'(0) = 4/4 = 1 > 0
hence f(x) is increasing at x = 0
graph{(-x^3+x^2-5x+6)/(x-2) [-10, 10, -5, 5]}