Question

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

Answer 1

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`