# How do you determine whether the function f(x)= (x-1) / (x+52) is concave up or concave down and its intervals?

Sep 17, 2015

Use the sign of the second derivative (or knowledge of transformations of the reciprocal function).

#### Explanation:

Calculus

Using calculus, the general method of determining concavity is to investigate the sign of the second derivative.

$f \left(x\right) = \frac{x - 1}{x + 52}$

$f ' \left(x\right) = \frac{53}{x + 52} ^ 2$

$f ' ' \left(x\right) = - \frac{106}{x + 52} ^ 3$

For this function, the sign of $f ' '$ is the opposite of the sign of $x + 52$.

$f ' '$ is positive on the interval $\left(- \infty , - 52\right)$ and negative on $\left(- 52 , \infty\right)$.

So the graph of $f$ is concave up interval $\left(- \infty , - 52\right)$ and concave down on $\left(- 52 , \infty\right)$.

Because $- 52$ is not in the domain of $f$, there is no inflection point.
(The definition of inflection point that I am accustomed to is: a point on the graph at which the concavity changes.)

Reciprocal Function

$f \left(x\right) = \frac{x - 1}{x + 52}$ can be written:

$f \left(x\right) = \frac{\left(x + 52\right) - 53}{x + 52} = \frac{x + 52}{x + 52} - \frac{53}{x + 52} = 1 - \frac{53}{x + 52}$

From the graph of $y = \frac{1}{x}$

graph{y=1/x [-20.28, 20.27, -10.14, 10.14]}

we obtain the graph of $f$ by
translating $52$ left, expanding vertically by a factor of $53$, reflect in the $x$ axis, and the translate up $1$ unit.

graph{y=(x-1)/(x+52) [-123.7, 42.94, -35.4, 48]}

Because of the reflection the graph is concave up on the left and concave down on the right. The horizontal translation moves the change in concavity from $x = 0$ to $x = - 52$