# How do you determine whether the function f(x)=(2x-3) / (x^2) is concave up or concave down and its intervals?

Jul 28, 2015

$f \left(x\right) = \frac{2 x - 3}{{x}^{2}}$ is concave up when $x > \frac{9}{2}$ and concave down when $x < \frac{9}{2}$ (and $x \ne 0$).

#### Explanation:

By the Quotient Rule, for $x \ne 0$, the first derivative is $f ' \left(x\right) = \frac{{x}^{2} \cdot 2 - \left(2 x - 3\right) \cdot 2 x}{{x}^{4}} = \frac{- 2 {x}^{2} + 6 x}{{x}^{4}} = \frac{- 2 x + 6}{{x}^{3}}$

and the second derivative is
$f ' ' \left(x\right) = \frac{{x}^{3} \cdot \left(- 2\right) - \left(- 2 x + 6\right) \cdot 3 {x}^{2}}{{x}^{6}} = \frac{4 {x}^{3} - 18 {x}^{2}}{{x}^{6}} = \frac{4 x - 18}{{x}^{4}}$

Since ${x}^{4} \setminus \ge q 0$ for all $x$, the sign of $f ' ' \left(x\right)$ is the same as the sign of its numerator $4 x - 18$. This expression is positive when $4 x > 18 \setminus \Leftrightarrow x > \frac{9}{2}$ and negative when $4 x < 18 \setminus \Leftrightarrow x < \frac{9}{2}$.

The value $x = \frac{9}{2}$ is the first coordinate of the unique inflection point of the graph of $f$. The second coordinate is $f \left(\frac{9}{2}\right) = \frac{9 - 3}{{\left(\frac{9}{2}\right)}^{2}} = \frac{6}{\frac{81}{4}} = \frac{24}{81} = \frac{8}{27}$