Is #f(x)=(x^2-3x-2)/(x+1)# increasing or decreasing at #x=1#?

1 Answer
Henry W.
Oct 8, 2016

Increasing

Explanation:

To determine if the graph is increasing or decreasing at a certain point, we can use the first derivative.

  • For values in which #f'(x)>0#, #f(x)# is increasing as the gradient is positive.
  • For values in which #f'(x)<0#, #f(x)# is decreasing as the gradient is negative.

Differentiating #f(x)#,

We have to use quotient rule.
#f'(x)=(u'v-v'u)/v^2#
Let #u=x^2-3x-2# and #v=x+1#
then #u'=2x-3# and #v'=1#

So #f'(x)=((2x-3)(x+1)-(x^2-3x-2))/(x+1)^2=(x^2+2x-1)/(x+1)^2#

Subbing in #x=1#,
#f'(x)=(1^2+2(1)-1)/(1+1)^2=1/2, :.f'(x)>0#

Since the #f'(x)>0# for #x=1#, #f(x)# is increasing at #x=1#

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