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f(x) = x2 + 1. For which values of x does f(x) =
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12 Jun 2017, 14:31

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Question Stats:

\(f(x) = x^2 + 1\). For which values of x does \(f(x) = f (\frac{1}{x})\)?

Indicate all such values:

❑ \(-2\)

❑ \(-1\)

❑ \(-\frac{1}{2}\)

❑ \(\frac{1}{2}\)

❑ 1

❑ 2

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f(x) = x2 + 1. For which values of x does f(x) =
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06 Jan 2022, 10:26

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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

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

Re: f(x) = x2 + 1. For which values of x does f(x) =
[#permalink]
24 Sep 2017, 08:47

It is quite straightforward to notice that the only value of x that are equal when considered as 1/x are -1 and +1. Since in the formula for f(x) the x is squared and then summed to 1, it is easy to see that every other solution would have lead to two different values when considering x and 1/x.

Answers B and E!

Answers B and E!

Re: f(x) = x2 + 1. For which values of x does f(x) =
[#permalink]
24 Sep 2017, 12:11

1

Carcass wrote:

\(f(x) = x^2 + 1\). For which values of x does \(f(x) = f (\frac{1}{x})\)?

Indicate all such values:

❑ -2

❑ -1

❑ \(\frac{-1}{2}\)

❑ 1/2

❑ 1

❑ 2

Show: :: OA

B and E

We have \(f(x) = f (\frac{1}{x})\) and given f(x) =\(x^2\)+1

now substituting we get -

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

or \(x^2\)+1 = \(\frac{1+x^2}{x^2}\)

or cross multiflication

\(x^4\)+\(x^2\) = 1+ \(x^2\)

or \(x^4\) =1

Only -1 and 1 fits the equation

Re: f(x) = x2 + 1. For which values of x does f(x) =
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27 Jul 2024, 07:24

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Re: f(x) = x2 + 1. For which values of x does f(x) = [#permalink]

27 Jul 2024, 07:24
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