Is $f (x) = \frac{3 x^{3} - 2 x^{2} - 2 x + 5}{x + 2}$ $f(x)=(3x^3-2x^2-2x+5)/(x+2)$ increasing or decreasing at $x = 3$ $x=3$ ?

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Answer 1 · 2016-01-07T06:13:05

$f'(x) =6x - 8 + 23/(x+2)^2$
and
$f'(3) = 273/25=10+23/25=10.92$

increasing

given

$f(x)= (3x^3 - 2x^2 -2x +5)/(x+2)$

proceed by dividing

$3x^3 - 2x^2 -2x + 5$ by $x+2$

to obtain

$f(x) = 3x^2 - 8x +14 -23/(x+2)$

find the first derivative to obtain

$f '(x) = 6x - 8+ 23/(x+2)^2$

evaluate

$f '(3) = 6(3)-8+23/(3+2)^2 = 10.92$

which indicates INCREASING at $x=3$

Is f(x)=(3x^3-2x^2-2x+5)/(x+2)f(x)=3x3−2x2−2x+5x+2f(x)=(3x^3-2x^2-2x+5)/(x+2) increasing or decreasing at x=3x=3x=3?