Question

Express the function in the form \(f \circ g \circ h\)

\(H\left( x \right) = {\sec ^4}\left( {\sqrt x } \right)\)

Short answer

The function \(H\left( x \right) = {\sec ^4}\left( {\sqrt x } \right)\)can be expressed in the form \(f \circ g\)with \(f\left( x \right) = {x^4}\) , \(g\left( x \right) = \sec x\) and \(h\left( x \right) = \sqrt x \).

Step-by-step solution

Given data

The provided function is

\(H\left( x \right) = {\sec ^4}\left( {\sqrt x } \right)\)

For three functions \(f\left( x \right)\), \(g\left( x \right)\)and\(h\left( x \right)\)their composition is defined as

\(f \circ g \circ h = f\left( {g\left( {h\left( x \right)} \right)} \right)\;\;\;\;\;.....\left( 1 \right)\)

Write the function

Let \(f\left( x \right) = {x^4}\), \(g\left( x \right) = \sec x\) and \(h\left( x \right) = \sqrt x \)

From equation (1),

\(\begin{array}{c}f \circ g \circ h = f\left( {g\left( {\sqrt x } \right)} \right)\\ = f\left( {\sec \left( {\sqrt x } \right)} \right)\\ = {\sec ^4}\left( {\sqrt x } \right)\\ = H\left( x \right)\end{array}\)

Hence, the required functions are \(f\left( x \right) = {x^4}\) , \(g\left( x \right) = \sec x\)and \(h\left( x \right) = \sqrt x \).