Question

1. find the domain 2. find vertical asymp 3. find horizontal asymp 4. find the critical points and intervals of increase and decrease of f(x) 5. Find the inflection points and intervals of concavity 6. Evaluate f(x) at rel max & min & at inflection point?

Answer 1

Explained below

Explanation:

Given function f(x)= `1/x`
1. This function is defined for all x on the number line, except at x=0. Hence the domain in interval form would be `(-oo,0)U(0,oo)`

2 &3.Vertical asymptote is x=0 or y- axis and horizontal asymptote is y=0 or x axis.

4.This function has no critical point, because at x=0 its derivative is undefined, but there are infinite number of points for which x-coordinate is 0.

In the interval `(-oo,0)` its derivative,f'(x) = `-1/x^2` will always be negative for all x. Samething applies for the interval `(0,oo)`. Hence it is a decreasing function in its entire domain

1. The second derivative of the function `1/x^3` which is not equal to 0 for any x-value of the function in its domain, hence it can be concluded that there is no inflection point.

2. Set the limit `x->oo or -oo` f(x)=0
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