Is $f (x) = \frac{- x^{3} + 2 x^{2} - 3 x + 7}{x - 2}$ $f(x)=(-x^3+2x^2-3x+7)/(x-2)$ increasing or decreasing at $x = - 1$ $x=-1$ ?

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Answer 1 · 2016-10-24T04:36:22

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

$y = \frac{f (x)}{g (x)}$ $y=f(x)/g(x)$

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

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.

Is f(x)=(-x^3+2x^2-3x+7)/(x-2)f(x)=−x3+2x2−3x+7x−2f(x)=(-x^3+2x^2-3x+7)/(x-2) increasing or decreasing at x=-1x=−1x=-1?