Question

Given f(x)=x^2+1, determine f(1/x), x does not equal to 0. Why is it:

Answer 1

well plug in 1/x to the function so thats

(1/x)^2 + 1

that equals (1/x^2) + 1

x cannot equal zero because 1/0 is undefined

Answer 2

Why is the process:

\[\left(\begin{matrix}1 \\ x ^{2}\end{matrix}\right) + \left(\begin{matrix}1\times x ^{2} \\ 1 \times x ^{2}\end{matrix}\right)\]

Answer 3

\[\left(\begin{matrix}1 \\ x ^{2}\end{matrix}\right) + \left(\begin{matrix}x ^{2} \\ x ^{2}\end{matrix}\right)\]

Answer 4

answer: \[\left(\begin{matrix}1 + x ^{2} \\ x ^{?}\end{matrix}\right)\]

Can someone explain to me why it's this process I should do?

Answer 5

x≠ 0 is just a condition to guarantee f(1/x) exist! You don't have to do anything.

Answer 6

that's what my teacher showed me but I'm having trouble understanding it.

Answer 7

x= 0 is not belong to domain of f(1/x)
Since if x= 0 --> f(1/x ) = ∞ ( undefined)

Answer 8

ohh, okay.

Answer 9

That's the rule, just memorize it whenever you see fraction, domain is numerator must be≠ 0

Answer 10

alright, thank you!

Answer 11

:)

Answer 12

I mean *denominator* must be ≠ 0!

Answer 13

okay!