Is #f(x)=(x^2-5x-9)/(2x+1)# increasing or decreasing at #x=0#?

Question

1448 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-15T23:10:49

At #x=0# the function is increasing

We must find the differential, hence allowing us to determine the gradient, and hence weather its increasing or decreasing...

Use the question rule..

#d/(dx) (u/v) = ( vdu-udv ) / (v^2) #

#=> d/dx ( (x^2-5x-9)/(2x+1) ) = ((2x+1)(2x-5) - (x^2-5x-9)(2) )/(2x+1)^2 #

When evaluating the differential at #x=0# we get:

#f'(0) = ((1*-5)-(-9*2))/(1^2) = 13 #

Hence the gradient is positive, #f'(0) > 0 # we can say the function is increasing at #x=0#

#y = (x^2-5x-9)/(2x+1) #

Desmos

As we can see by the graph it may be negative at #x=0# by the gradient is positive, it is increasing...