f(x) = x^2 + 1. For which values of x does f(x) = f(1/x)?

Question

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

I. -1

II.  -\frac{1}{2}

III.  \frac{1}{2}

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Archit3110 · Answer

IMO A ; for f(-1) = f(1/-1) ;;  f(x) = x^2 + 1  = 2 
will be valid

nick1816 · Answer

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

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

ScottTargetTestPrep · Answer

Let’s check each Roman numeral.

I. -1

f(-1) = (-1)^2 + 1 = 2

f(1/(-1)) = f(-1) = 2

I is true.

II. -½

f(-½) = (-½)^2 + 1 = ¼ + 1 = 5/4

f(1/(-1/2)) = f(-2) = (-2)^2 + 1 = 4 + 1 = 5

II is not true.

II. ½

f(½) = (½)^2 + 1 = ¼ + 1 = 5/4

f(1/(1/2)) = f(2) = 2^2 + 1 = 4 + 1 = 5

III is not true.

Answer: A

Pradd1984 · Answer

Can someone please why substituting (f)x x^2+1 into f(1/x) 1/x is not 1/(x^2 + 1) and why it is instead 1/(x^2) + 1 ?

Innes · Answer

I was getting confused in this as well but then I figured that I can substitute values and then check the form:

f(2) = (2)^2 + 1 
f(1/2) = (1/2)^2+1

Hope this helps.

BrushMyQuant · Answer

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} )

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

II  -\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)  
eq  f( \frac{1}{x} ) FALSE

III  \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)  
eq  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 A 
Hope it helps!

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

https://www.youtube.com/watch?v=b7JoMX0gk8M

f(x) = x^2 + 1. For which values of x does f(x) = f(1/x)?

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f(x) = x^2 + 1. For which values of x does f(x) = f(1/x)?

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