If $f (\frac{x}{x + 1}) = {(x - 1)}^{2}$ $f(x/(x+1))=(x-1)^2$ , what is $f (3)$ $f(3)$ ?

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Answer 1 · 2018-06-04T18:40:58

$\frac{25}{4} = 6.25$ $25/4=6.25$ .

Given that, $f (\frac{x}{x + 1}) = {(x - 1)}^{2}$ $f(x/(x+1))=(x-1)^2$ .

We require the value of $f (3)$ $f(3)$ .

Clearly the corresponding $x$ $x$ for this can be obtained by solving

$\frac{x}{x + 1} = 3$ $x/(x+1)=3$ .

$:. x=3x+3$ .

$:. -2x=3," giving, "x=-3/2$ .

Subst.ing $x=-3/2$ in the formula for $f$ , we get,

$f((-3/2)/(-3/2+1))=f(-3/2-:-1/2)=f(3)=(-3/2-1)^2$ .

$rArr f(3)=25/4=6.25$ .

Enjoy Maths.!

Answer 2 · 2018-06-04T18:47:11

$f(3) = 25/4$

We are looking for what value of $x$ gives

$x/(x+ 1) = 3$

Solve the equation:

$x = 3x + 3$

$-2x = 3$

$x =-3/2$

Now simply plug $x= -3/2$ into the formula

$f(3) = (-3/2 - 1)^2 = 25/4$

If f(x/(x+1))=(x-1)^2f(xx+1)=(x−1)2f(x/(x+1))=(x-1)^2, what is f(3)f(3)f(3)?