# Vertical asymptotes, intervals of increase and decrease, local maxima and minima, concavity, inflection for $f(x)=1+(7/x)-(5/x^2)$

$$f(x) = 1 + \frac {7}{x} - \frac {5}{x^2}$$

(a) Find the vertical asymptotes.

I answered $1$ because if you plug in $0$ for $x$ you get $1$.

(b) Find the interval where the function is increasing.

I answered ($\frac {7}{10}, \infty$) because I got $f'(x) = \frac {1}{x}(\frac{-7}{x} + 10)$ and solved for x.

(c) Find the interval where the function is decreasing.

I answered ($-\infty, \frac {7}{10}$)

(d) Find the local minimum and maximum values.

I answered local maximum $=$ DNE and local minimum $\approx$ .7959.

(e) Find the interval where the function is concave up.

I answered ($-\infty, \frac{7}{5}$) because I found $f''(x) = \frac{1}{x^2}(\frac{14}{x} - 10)$. Then I solved for $x$.

(f) Find the interval where the function is concave down.

I answered ($\frac{7}{5}, \infty$)

(g) Find the inflection point.

I answered ($\frac{7}{5}, 3.45$)

For (a), if you plug in $x=0$, you don't get $1$ --- you don't get anything, since you can't plug in $x=0$, since division by zero is undefined. In fact, it's for that reason that $x=0$ is the vertical asymptote.