Is f(x)=(x^3+7x^2-5x-4)/(x-2) increasing or decreasing at x=0?

Mar 14, 2018

The function $f \left(x\right)$ is increasing at $x = 0$.

Explanation:

To find out if the function is increasing or decreasing at $x = 0$, take the derivative, and plug in $x = 0$ and see if it's positive or negative.

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\frac{{x}^{3} + 7 {x}^{2} - 5 x - 4}{x - 2}\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{\frac{d}{\mathrm{dx}} \left({x}^{3} + 7 {x}^{2} - 5 x - 4\right) \cdot \left(x - 2\right) - \left({x}^{3} + 7 {x}^{2} - 5 x - 4\right) \cdot \frac{d}{\mathrm{dx}} \left(x - 2\right)}{x - 2} ^ 2$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{\left(3 {x}^{2} + 14 x - 5\right) \left(x - 2\right) - \left({x}^{3} + 7 {x}^{2} - 5 x - 4\right) \cdot 1}{x - 2} ^ 2$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{\left(3 {x}^{2} + 14 x - 5\right) \left(x - 2\right) - {x}^{3} - 7 {x}^{2} + 5 x + 4}{x - 2} ^ 2$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{3 {x}^{3} - 6 {x}^{2} + 14 {x}^{2} - 28 x - 5 x + 10 - {x}^{3} - 7 {x}^{2} + 5 x + 4}{x - 2} ^ 2$

$\textcolor{w h i t e}{f ' \left(x\right)} = \frac{2 {x}^{3} + {x}^{2} - 28 x + 14}{x - 2} ^ 2$

Now, let's plug $x = 0$. If the result is positive, then the function is increasing at $x = 0$. If it's negative, then it's decreasing.

$\textcolor{w h i t e}{\implies} \frac{2 {x}^{3} + {x}^{2} - 28 x + 14}{x - 2} ^ 2$

$\implies \frac{2 {\left(0\right)}^{3} + {\left(0\right)}^{2} - 28 \left(0\right) + 14}{0 - 2} ^ 2$

$= \frac{0 + 0 - 0 + 14}{- 2} ^ 2$

$= \frac{14}{4}$

$= \frac{7}{2}$

Since this value is positive, the original function $f \left(x\right)$ is increasing at $x = 0$.

We can verify this by looking at the graph of $f \left(x\right)$:

graph{(((x^3+7x^2-5x-4)/(x-2))) [-3, 3, -10, 6]}