Question about Concavity?

Question

How do I find all the values on which the graph of the function f(x)= (x^2+1)/x^2 is concave upwards?

Answer 1

1) Differentiate twice

a) f'(x) = -2/(x ^ 3)
b) f"(x) = 6/(x ^ 4)

2) Find the inflection points by

(a) setting f"(x) = 0 and by
(b) finding where f"(x) is undefined.

0 = 6/(x ^ 4), but this is never equal to 0.

6/(x ^ 4) is undefined when x = 0, which means that x = 0 is an inflection point

3) Test points on either side of that/those point(s).

Simple points to try are x = -1 and x = 1.

The test point(s) is/are plugged into f"(x).
a) If the resulting value was positive, then the values on that side of the appropriate inflection point would be concave up.

b) If the resulting value was negative, then the values on that side of the appropriate inflection point would be concave down.

When x = -1, f"(-1) = 6.
When x = 1, f"(1) = 6.

Both resulting values are positive. This means that on both sides of the inflection point x = 0, the graph of f(x) is concave up, thus f(x) is concave up on all reals (but not 0).

Answer 2

First, do you know what that means? That's the first step to solving problems: understanding definitions.

Look up the definition of concave upwards then come back and ask for help.

You can also email me.

Answer 3

Differentiate twice. Where f''(x), or acceleration is positive, the function f(x) concaves upwards.