Question

If the value of $f(x) = {(25 - {x^4})^{1/4}}$ , $0 < x < \sqrt 5$ then $f\left( {f\left( {\dfrac{1}{2}} \right)} \right) =$
A) ${2^{ - 4}}$
B) ${2^{ - 3}}$
C) ${2^{ - 2}}$
D) ${2^{ - 1}}$

Answer 1

Here, we have to evaluate for the given function. A function is a process or a relation that associates each element $x$ of a set $X$, the domain of the function, to a single element y of another set Y (possibly the same set), the co-domain of the function. We will first solve the given function and find the value of $f(f(x))$. Then substitute the value in the required function to find the value.
Formula used:
Exponential Formula: When ${a^m}$ is raised to the power $n$, we have ${({a^m})^n} = {a^{m \cdot n}}$.

Complete step by step solution:
We are given with a function $f(x) = {(25 - {x^4})^{\dfrac{1}{4}}}$.
$x$ should lie between $0$ and $\sqrt 5$.
$\Rightarrow f(x) = {\left( {25 - {x^4}} \right)^{\dfrac{1}{4}}}$
Now, the function of $f(f(x))$ is given by
$\Rightarrow f\left( {f(x)} \right) = {\left( {25 - {{\left( {{{\left( {25 - {x^4}} \right)}^{\dfrac{1}{4}}}} \right)}^4}} \right)^{\dfrac{1}{4}}}$
When ${a^m}$ is raised to the power $n$, we have ${({a^m})^n} = {a^{mn}}$
Subtracting the terms, we have
$\Rightarrow f\left( {f\left( x \right)} \right) = {\left( {25 - 25 + {x^4}} \right)^{\dfrac{1}{4}}}$
When ${a^m}$ is raised to the power $n$, we have ${({a^m})^n} = {a^{mn}}$
$\Rightarrow f\left( {f\left( x \right)} \right) = {({x^4})^{\dfrac{1}{4}}}$
$\Rightarrow f\left( {f\left( x \right)} \right) = x$
Substituting the value of $x = \dfrac{1}{2}$ , we have
$\Rightarrow f\left( {f\left( {\dfrac{1}{2}} \right)} \right) = \dfrac{1}{2}$
$\Rightarrow f\left( {f\left( {\dfrac{1}{2}} \right)} \right) = {2^{ - 1}}$

Therefore, $f\left( {f\left( {\dfrac{1}{2}} \right)} \right) = {2^{ - 1}}$

Note:
We should know about the relation and a function. A relation is a set of inputs and outputs, and a function is a relation with one output for each input. A function is a relation which describes that there should be only one output for each input (or) we can say that a special kind of relation (a set of ordered pairs), which follows a rule i.e. every $X$-value should be associated with only one $y$-value is called a function. The relation between the two sets is defined as the collection of the ordered pair, in which the ordered pair is formed by the object from each set. Function may be classified into various types of functions based on the relation. All functions are relations, but not all relations are functions.