Is $f (x) = \frac{2 x^{3} - 7 x^{2} + 13 x - 11}{x - 3}$ $f(x)=(2x^3-7x^2+13x-11)/(x-3)$ increasing or decreasing at $x = 2$ $x=2$ ?

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Answer 1 · 2018-07-15T12:31:49

$f (x)$ $f(x)$ is monotonously decreasing in the point $x = 2$ $x=2$

$(u/v)'=(u'v-uv')/v^2$
By this rule we get

$f'(x)=((6x^2-14x+13)(x-2)-(2x^3-7x^2+13x-11))/(x-3)^2$

expanding the numerator

$f'(x)=(6x^3-14x^2+13x-18x^2+42x-39-2x^3+7x^2-13x+11)/(x-3)^2$

collecting like Terms

$f'(x)=(4x^3-25x^2+42x-28)/(x-3)^2$

$f'(2)=(32-100+84-28)/(-1)^2=-12<0$

Is f(x)=(2x^3-7x^2+13x-11)/(x-3)f(x)=2x3−7x2+13x−11x−3f(x)=(2x^3-7x^2+13x-11)/(x-3) increasing or decreasing at x=2x=2x=2?