Is f(x)=(-x^2+5x-2)/(x^2-1) increasing or decreasing at x=-3?

Mar 26, 2016

$f \left(x\right) = \frac{- {x}^{2} + 5 x - 2}{{x}^{2} - 1}$ is always decreasing for all values of $x$ including at $x = - 3$

Explanation:

To find whether $f \left(x\right) = \frac{- {x}^{2} + 5 x - 2}{{x}^{2} - 1}$ is increasing or decreasing at $x = - 3$, we should find the value of $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}$ at $x = 3$.

$\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}} = \frac{\left({x}^{2} - 1\right) \left(- 2 x + 5\right) - \left(- {x}^{2} + 5 x - 2\right) \left(2 x\right)}{{x}^{2} - 1} ^ 2$ or

= $\frac{\left(- 2 {x}^{3} + 2 x + 5 {x}^{2} - 5\right) - \left(- 2 {x}^{3} + 10 {x}^{2} - 4 x\right)}{{x}^{2} - 1} ^ 2$ or

= $\frac{- 5 {x}^{2} + 6 x - 5}{{x}^{2} - 1} ^ 2$ and at $x = - 3$, we have

$f ' \left(3\right) = \frac{- 5 \times {\left(- 3\right)}^{2} + 6 \times \left(- 3\right) - 5}{{\left(- 3\right)}^{2} - 1} ^ 2$

= $\frac{- 45 - 18 - 5}{64} = - \frac{68}{64}$

As $\frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}$ is negative at $x = - 3$

$f \left(x\right) = \frac{- {x}^{2} + 5 x - 2}{{x}^{2} - 1}$ is decreasing.

Note that in $\left(- 5 {x}^{2} + 6 x - 5\right)$

= $- 5 \left({x}^{2} - \left(\frac{6}{5}\right) x + {\left(\frac{3}{5}\right)}^{2} - {\left(\frac{3}{5}\right)}^{2} + 1\right)$ or

= $- 5 \left\{{\left(x - \frac{3}{5}\right)}^{2} + \frac{16}{25}\right\}$ and hence negative for all $x$

Hence it is always decreasing.

graph{(-x^2+5x-2)/(x^2-1) [-10, 10, -5, 5]}