For what values of x is $f (x) = \frac{1}{x - 3}$ $f(x)=1/(x-3)$ concave or convex?

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Answer 1 · 2016-11-22T00:50:33

Concave UP on the interval ( $- \infty, 3$ $-oo, 3$ )
Concave DOWN on the interval ( $3, \infty$ $3, oo$ )

Take the first derivative of f(x)

$f'(x) = -1/(x-3)^2$

Then second derivative:

$f''(x) = 2/(x-3)^3$

Find the points where $f''(x) = 0$ or is undefined.

$f(x)$ is undefined when $x=3$

Plug in a number $x>3$ and you will see the function will be positive; thus the function is concave up.

Since the multiplicity of the function $(x-3)^3$ is an odd function, the values for $x<3$ will be negative; thus the function will be concave down.

For what values of x is f(x)=1/(x-3)f(x)=1x−3f(x)=1/(x-3) concave or convex?