Question

Determine the intervals where the graph of the function f(x)=x+1/x is concave up and concave down and inflection point?

Answer 1

concave up: `(0, oo)`; concave down: `(-oo, 0)`
no inflection point

Explanation:

Given: `f(x) = x + 1/x = (x^2 + 1)/x`

There is a vertical asymptote at `x = 0`

Find the 1st derivative:
`f'(x) = 1 = x^-2 = 1 - 1/x^2`

Find the critical values: `f'(x) = 0`
`1 - 1/x^2 = 0`;

`1 = 1/x^2`

`x^2 = 1; " "x = +- 1`

Find Critical points:
`f(-1) = -2; " "f(1) = 2; " critical points:"(-1, -2), (1, 2)`

Find the inflection points `f''(x) = 0`
`f''(x) = 2x^-3 = 2/x^3 = 0; " " x = 0`

`f(0) = "undefined"`
No inflection points.

Use the first derivative test to find the concave, convex intervals:

Intervals: `(-oo, -1), " "(-1, 0), " "(0, 1), " "(1, oo)`
Test value: `x = -2; " "x = -1/2; " "x = 1,2; " "x = 2`
`f'(-2) > 0; " "f'(-1/2)< 0; " "f'(1/2) < 0; " "f'(2)>0`

Relative max: `(-1, -2)`

asymptote at `x = 0`

Relative min.: `(1, 2)`

concave up: `(0, oo)`; concave down: `(-oo, 0)`

graph{x+1/x [-10, 10, -5, 5]}