Function with Exponential

Observe: $\frac{16+y^2}{y^2}=(\frac{4}{y})^2+1$ therefore it does not take too long for us to realise that we must make the substitution $x=\frac{4}{y}$ and by doing this we get:

$f(\frac{16+y^2}{y^2})^{\frac{2}{\sqrt{y}}}=16$ or

$(f(\frac{16+y^2}{y^2})^{\frac{1}{\sqrt{y}}})^2=16$ and therefore

$f(\frac{16+y^2}{y^2})^{\frac{1}{\sqrt{y}}}=\sqrt{16}=4$

First note that $\large \frac{16+y^2}{y^2}=\frac{16}{y^2}+1= \left(\frac{4}{y}\right)^2+1$ . Then letting $\large x=\frac{4}{y}$ we get:

$\large f \left(\left(\frac{4}{y}\right)^2+1\right)^\frac{2}{\sqrt{y}}=16$

Raising both sides to the exponent $\large \frac{1}{2}$ gives us:

$\large \sqrt{f \left(\left(\frac{4}{y}\right)^2+1\right)^\frac{2}{\sqrt{y}}}=f \left(\left(\frac{4}{y}\right)^2+1\right)^\frac{1}{\sqrt{y}}=\sqrt{16}=\pm 4$

We take the principal root and so our answer is $\boxed{4}$ .

First of all I'm going to do some algebra, $\frac{16+y^2}{y^2} =\frac{16}{y^2}+1=(\frac{4}{y})^2 +1.$ On the other hand $f(x^2 +1)^{\sqrt{x}}=16\\ f(x^2 +1)=(16)^{\frac{1}{\sqrt{x}}}.$ Define $x=\frac{4}{y},$ and substituting x

$f((\frac{4}{y})^2 + 1)=16^{\frac{1}{\sqrt{\frac{4}{y}}}} \\ f((\frac{4}{y})^2 + 1)=16^{\frac{\sqrt{y}}{2}}\\ f((\frac{4}{y})^2 + 1)=16^{\frac{1}{2}\sqrt{y}} \\ f((\frac{4}{y})^2 + 1)=\sqrt{16}^{\sqrt{y}} \\ f((\frac{4}{y})^2 + 1)=4^{\sqrt{y}} \\ f((\frac{4}{y})^2 + 1)^{\frac{1}{\sqrt{y}}}=4.$ Hence $f( \frac{16+y^2}{y^2} )^{\frac{1}{\sqrt{y}}}=4$

Simplify $f(\frac{16+y^{2}}{y^{2}})^{\frac{1}{\sqrt{y}}}$ we get...

$f(\frac{16+y^{2}}{y^{2}})^{\frac{1}{\sqrt{y}}}$

$=f(\frac{16}{y^{2}}+1)^{\frac{1}{\sqrt{y}}}$

Substitute $x = \frac{4}{y}$ -> $\frac{1}{\sqrt{y}} = \frac{\sqrt{x}}{2}$ -> $\frac{16}{y^{2}} = x^{2}$

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

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

$=\sqrt{16} = \boxed{4}$ ~~~

Well, first, all the information we have about the function is the relation that is shown in the problem. So, for avoiding further complication, we will require that the argument of $f$ to be equal on both cases, i.e.,

$x^{2} + 1 = \frac{16 + y^{2}}{y^{2}} \therefore \frac{1}{\sqrt{y}} = \frac{\sqrt{x}}{2}$

Now it is easy to see that

$\left [ f \left ( \frac{16 + y^{2}}{y^{2}} \right ) \right ]^{\frac{1}{\sqrt{y}}} = \left [ f \left ( x^{2} + 1 \right ) \right ]^{\frac{\sqrt{x}}{2}} = \sqrt{16} = 4$

Cheap : If $x = 4$ , we have

$f \left( 17 \right)^{\sqrt{4}} = f \left( 17 \right)^2 = 16 \implies f(17) = \pm 4.$

Since $f$ is a function on the positive reals, $f(17) = 4$ .

Now, if $y = 1$ , we have:

$f \left( \dfrac {16+y^2}{y^2} \right)^{\frac{1}{\sqrt{y}}} = f \left( 17 \right)^1 = f(17) = \boxed {4}.$

In order to solve this problem, we should try to get the function $f(\frac{16+y^{2}}{y^{2}})^{\frac{1}{\sqrt{y}}}$ to look like something we know: $f(x^{2}+1)^{\sqrt{x}}$ .
With some algebraic manipulation we find that: $f(\frac{16+y^{2}}{y^{2}})^{\frac{1}{\sqrt{y}}} = f(\frac{16}{y^{2}} + 1)^{\frac{1}{\sqrt{y}}} = f((\frac{4}{y})^{2}+1)^{\frac{2}{\sqrt{y}} \times \frac{1}{2}}$ .
Now, if we set $x=\frac{4}{y}$ , we find that $f((\frac{4}{y})^{2}+1)^{\frac{2}{\sqrt{y}} \times \frac{1}{2}} = (f(x^{2}+1)^{\sqrt{x}})^{\frac{1}{2}}$
Since we know $f(x^{2}+1)^{\sqrt{x}} = 16$ , the $(f(x^{2}+1)^{\sqrt{x}})^{\frac{1}{2}}$ is just $16^{\frac{1}{2}} = \sqrt{16} = \boxed{4}$

The function $f$ is given such that $[f(x^2+1)]^{\sqrt{x}}=16$ and $f$ is a function on the positive reals. Then, we have --

$[f(\frac{16+y^2}{y^2})]^{\frac{1}{\sqrt{y}}}$

$=[f(\frac{16}{y^2}+1)]^{\frac{1}{\sqrt{y}}}$

$=[f((\frac{4}{y})^2+1)]^{\frac{2}{\sqrt{y}}\times \frac{1}{2}}$

$=([f((\frac{4}{y})^2+1)]^{\frac{2}{\sqrt{y}}})^{\frac{1}{2}}$

$=(16)^{\frac{1}{2}}$ [since, if we replace $x$ in $[f(x^2+1)]^{\sqrt{x}}=16$ with $\frac{4}{y}$ , then we get $([f((\frac{4}{y})^2+1)]^{\frac{2}{\sqrt{y}}})=16$ ]

$=\boxed{4}$ [Neglecting (-4) as the function is defined on (+ve) reals]

Use $(4/y) = z$ Given LHS reduces to $[f(z^{2} + 1)]^{\sqrt{z}/2} = (16)^{1/2} =( 2^{4})^{1/2} = 4$

Take $x=\frac{4}{y}$ and from this $\left( f((16+y^2)/(y^2)) \right)^{\frac{1}{\sqrt{y}}}=f(x^2+1)^{\sqrt{x}/2} = 4$

Let $x^{2} = \frac {16}{y^{2}}$ . $\Rightarrow [f(\frac {16}{y^{2}} + 1)]^{\frac {2}{\sqrt{y}}} = [f(\frac {16 + y^{2}}{y^{2}})]^{\frac {2}{\sqrt{y}}} = 16$

Then, taking the square root of both sides yields: $\boxed{[f(\frac {16 + y^{2}}{y^{2}})]^{\frac {1}{\sqrt{y}}} = 4}$

let x =x\4 the the above equation become f{(16+x x)/x x} to the power 2/xto the power 1\2 that will give y=4

We can rewrite the given argument as:

$\dfrac{16+y^{2}}{y^{2}} = \dfrac{16}{y^{2}} + 1 = \left(\dfrac{4}{y}\right)^{2} + 1$

This is the same structure as the argument of the given function, where $x = \frac{4}{y}$ !

Then:

$\begin{aligned}f \left( x^{2} + 1 \right) ^ {\sqrt{x}} &= f\left(\left(\dfrac{4}{y}\right)^{2} + 1\right)^{\sqrt{\dfrac{4}{y}}}\\ &= f\left(\left(\dfrac{4}{y}\right)^{2} + 1\right)^{\dfrac{2}{\sqrt{y}}}\end{aligned}$

This above expression is the square of the required value (look at the exponent). But this expression is equal to $f \left( x^{2} + 1 \right) ^ {\sqrt{x}} = 16$ . Hence,

$f\left(\dfrac{16+y^{2}}{y^{2}}\right)^{\dfrac{1}{\sqrt{y}}} = \sqrt{16} = \boxed{4}$