Given that \(f(x) = x^2 + 1\) and we need to find for which values of x does \(f(x) = f (\frac{1}{x})\)

To find \(f(\frac{1}{x})\) we need to compare what is in the bracket in \(f(\frac{1}{x})\) and what is in the bracket in f(x)

=> to find the value of \(f(\frac{1}{x})\) we need to replace x with \(\frac{1}{x}\) in f(x)
=> \(f(\frac{1}{x})\) = \((\frac{1}{x})^2 + 1\) = \(\frac{1 + x^2 }{ x^2}\)

Now, there are two ways to solve this:

Method 1: Substitution

Take each value and find value of f(x) and f(\(\frac{1}{x}\)) and see for which value is f(x) = f(\(\frac{1}{x}\))

A \(-2\)
f(x) = f(-2) = \((-2)^2 + 1\) = 5
f(\(\frac{1}{x}\)) = f(\(\frac{1}{-2}\)) = \(\frac{1 + (-2)^2 }{ (-2)^2}\) = \(\frac{5}{4}\)
f(x) \(\neq\) f(\(\frac{1}{x}\)) FALSE

B \(-1\)
f(x) = f(-1) = \((-1)^2 + 1\) = 2
f(\(\frac{1}{x}\)) = f(\(\frac{1}{-1}\)) = \(\frac{1 + (-1)^2 }{ (-1)^2}\) = \(\frac{2}{1}\) = 2
f(x) =f(\(\frac{1}{x}\)) TRUE

C \(-\frac{1}{2}\)
f(x) = f(\(\frac{-1}{2}\)) = \((\frac{-1}{2})^2 + 1\) = \(\frac{5}{4}\)
f(\(\frac{1}{x}\)) = f(\(1/(\frac{-1}{2})\)) = f(-2) = \((-2)^2 + 1\) = 5
f(x) \(\neq\) f(\(\frac{1}{x}\)) FALSE

D \(\frac{1}{2}\)
f(x) = f(\(\frac{1}{2}\)) = \((\frac{1}{2})^2 + 1\) = \(\frac{5}{4}\)
f(\(\frac{1}{x}\)) = f(\(1/(\frac{1}{2})\)) = f(2) = \((2)^2 + 1\) = 5
f(x) \(\neq\) f(\(\frac{1}{x}\)) FALSE

E 1
f(x) = f(1) = \((1)^2 + 1\) = 2
f(\(\frac{1}{x}\)) = f(\(\frac{1}{1}\)) = \(\frac{1 + (1)^2 }{ (1)^2}\) = \(\frac{2}{1}\) = 2
f(x) =f(\(\frac{1}{x}\)) TRUE

F 2
f(x) = f(2) = \((2)^2 + 1\) = 5
f(\(\frac{1}{x}\)) = f(\(\frac{1}{2}\)) = \(\frac{1 + (2)^2 }{ (2)^2}\) = \(\frac{5}{4}\)
f(x) \(\neq\) f(\(\frac{1}{x}\)) FALSE

Method 2: Algebra

f(x) = \(f(\frac{1}{x})\)
=> \(x^2 + 1 = \frac{1 + x^2 }{ x^2}\)
=> \(x^2*(x^2 + 1) = 1 + x^2 \)
=> \(x^4 + x^2 - 1 - x^2 = 0\)
=> \(x^4 = 1\)
=> x = +1 or -1

So, Answer will be B and E
Hope it helps!

Watch the following video to learn the Basics of Functions and Custom Characters