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F(x + 1/x) = (x + 1/x)^2 - 2
So, the function can easily be reduced to F(y) = y^2 - 2 where y = x + 1/x
Therefore, F(4) + F(5) = (4^2-2) + (5^2-2) = 14 + 23 = 37
Pick D
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keats
F(x + 1/x) = (x + 1/x)^2 - 2
So, the function can easily be reduced to F(y) = y^2 - 2 where y = x + 1/x
Therefore, F(4) + F(5) = (4^2-2) + (5^2-2) = 14 + 23 = 37
Pick D

Can someone please provide a better explanation?
Thanks in advance!
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pushpitkc
We know that the function
F(x+\(\frac{1}{x}\))=\(x^2 + \frac{1}{x^2}\)

To find F(4)
\(x + \frac{1}{x}\) = 4
Squaring on both sides
\(x^2 + \frac{1}{x^2} + 2*x^2*\frac{1}{x^2}\) = 16
\(x^2 + \frac{1}{x^2}\) = 16-2 =14

To find F(5)
\(x + \frac{1}{x}\) = 5
Squaring on both sides
\(x^2 + \frac{1}{x^2} + 2*x^2*\frac{1}{x^2}\) = 25
\(x^2 + \frac{1}{x^2}\) = 25-2 =23

So the sum of F(4) and F(5) = 14+23 = 37(Option D)

Could you please elaborate more on this part:

Squaring on both sides
\(x^2 + \frac{1}{x^2} + 2*x^2*\frac{1}{x^2}\) = 16
\(x^2 + \frac{1}{x^2}\) = 16-2 =14

It is not clear why and how you square both sides where digit 2 comes from?
Thank you very much!
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\((a+b)^2 = a^2 + b^2 + 2*a*b\)
if a = x and b = 1/x

\((x+\frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2*x*\frac{1}{x}\)
Since x and \frac{1}{x} cancel out each other, we get 2.
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pushpitkc
\((a+b)^2 = a^2 + b^2 + 2*a*b\)
if a = x and b = 1/x

\((x+\frac{1}{x})^2 = x^2 + \frac{1}{x^2} + 2*x*\frac{1}{x}\)
Since x and \frac{1}{x} cancel out each other, we get 2.

Thank you so much! Greatly appreciated!
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If \(F(x + \frac{1}{x}) = x^2 + \frac{1}{x^2}\), what is the value of \(F(4) + F(5)\).


    A. 9
    B. 16
    C. 25
    D. 37
    E. 41


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If we square \(x + \frac{1}{x}\) we will get \(x^2 + \frac{1}{x^2} + 2\).
So we can write \(F(x + \frac{1}{x})\) = \((x +\frac{1}{x})^2\) - 2 = \(x^2 + \frac{1}{x^2}\)
---> F(4) = (4)^2 - 2 = 14
and F(5) = (5)^2 - 2 = 23
Adding, 37.
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Ans is D

f(x+ 1/x)= x^2 + 1/x^2 = ( x+1/x)^2-2
let x+1/x = y
f(y) = y^2-2
f(4)=16-2
f(5)=25-2
f(4)+f(5)=16+25-2-2=37
Option D
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F(x+1/X) = (x+1/x)^2 - 2
now put x+1/x = 4
F(4) = 16 - 2=14
same way F(5) = 23
total = 37
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If \(F(x + \frac{1}{x}) = x^2 + \frac{1}{x^2}\), what is the value of \(F(4) + F(5)\)?

Looking at the information provided, we can recognize the following.

If we square \((x + \frac{1}{x})\), we get \((x + \frac{1}{x})^2 = x^2 + 2 + \frac{1}{x^2}\).

So, \(x^2 + \frac{1}{x^2} = (x + \frac{1}{x})^2 - 2\).

Thus, if \(F(x + \frac{1}{x}) = x^2 + \frac{1}{x^2}\), then \(F(x) = x^2 - 2\).

\(F(4) + F(5) = (4^2 - 2) + (5^2 - 2) = 14 + 23 = 37\)

A. 9
B. 16
C. 25
D. 37
E. 41


Correct answer: D
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