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

##### 1 Answer
Jan 28, 2018

The function is increasing at $x = 2$. See explanation

#### Explanation:

To find out if a function $f \left(x\right)$ is increasing or decreasing at a given point ${x}_{0}$ you can calculate the first derivative $f ' \left({x}_{0}\right)$.

If the derivative $f ' \left({x}_{0}\right)$ is greater than zero, then the function is increasing, if it is negative, then the function is decreasing.

Here the derivative can be calculated using the quotient rule:

$f \left(x\right) = \frac{2 {x}^{3} - 7 {x}^{2} + 3 x + 1}{x - 3}$

$f ' \left(x\right) = \frac{\left(2 {x}^{3} - 7 {x}^{2} + 3 x + 1\right) ' \left(x - 3\right) - \left(2 {x}^{3} - 7 {x}^{2} + 3 x + 1\right) \left(x - 3\right) '}{x - 3} ^ 2$

$f ' \left(x\right) = \frac{\left(6 {x}^{2} - 14 x + 3\right) \left(x - 3\right) - \left(2 {x}^{3} - 7 {x}^{2} + 3 x + 1\right)}{x - 3} ^ 2$

$f ' \left(x\right) = \frac{6 {x}^{3} - 32 {x}^{2} + 45 x - 9 - 2 {x}^{3} + 7 {x}^{2} - 3 x - 1}{x - 3} ^ 2$

$f ' \left(x\right) = \frac{4 {x}^{3} - 25 {x}^{2} + 42 x - 10}{x - 3} ^ 2$

Now we calculate the value of $f ' \left(2\right)$

$f ' \left(2\right) = \frac{4 \cdot {2}^{3} - 25 \cdot {2}^{2} + 42 \cdot 2 - 10}{2 - 3} ^ 2$

$f ' \left(2\right) = 32 - 100 + 84 - 10 = 6$

$f ' \left(2\right)$ is greater than zero, so $f \left(x\right)$ is increasing at $x = 2$