Question 1088254
<font color="black" face="times" size="3">We have {{{f(sqrt(x+1))}}} and we want to find {{{f(2)}}}. The inner terms {{{sqrt(x+1)}}} and {{{2}}} must be equal. So we have


{{{sqrt(x+1) = 2}}}


which solves to


{{{sqrt(x+1) = 2}}}
{{{(sqrt(x+1))^2 = 2^2}}}
{{{x+1 = 4}}}
{{{x+1-1 = 4-1}}}
{{{x = 3}}}


Now let's plug x = 3 into the function below


{{{f(sqrt(x+1)) = 1/x}}}
{{{f(sqrt(3+1)) = 1/3}}}
{{{f(sqrt(4)) = 1/3}}}
{{{f(2) = 1/3}}}
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