Question

Is `f(x)=(-x^3+2x^2-3x+7)/(x-2)` increasing or decreasing at `x=-1`?

Answer 1

So first we have to find the derivative of the function,

`y=f(x)/g(x)`

`y'=(f'(x)g(x)-g'(x)f(x))/g(x)^2`

so with,
`y=(-x^3+2x^2-3x+7)/(x-2)`

`f(x)=-x^3+2x^2-3x+7`

`f'(x)=-3x^2+4x-3`

and,

`g(x)=x-2`

`g'(x) = 1`

so subbing into the formula,

`y'=((-3x^2+4x-3)(x-2)-(-x^3+2x^2-3x+7))/(x-2)^2`

`y'=(-2x^3-8x^2+8x+1)/(x-2)^2`

now to find what the gradient is at `x=-1`
we need to sub in `x=-1` into `y'`

`y'=(-2(-1)^3-8(-1)^2+8(-1)+1)/((-1)-2)^2`

`y'=17/9`

this means that the gradient is positive at `x=-1` and as a result the function in increasing at this value.