Question

(a) If\(g(x) = 2x + 1 {\rm{and}} h(x) = 4{x^2} + 4x + 7\), find a function f such that\(f \circ g = h\). (Think about what operations you would have to perform on the formula for g to end up with the formula for h.)

(b) If\(f(x) = 3x + 5 {\rm{and}} h(x) = 3{x^2} + 3x + 2\), find a function g such that\(f \circ g = h\).

Short answer

(a)\(f(x) = {x^2} + 6\)

(b)\(g(x) = \left( {{x^2} + x - 1} \right)\)

Step-by-step solution

Introduction

Two functions can be expressed as combination of each other with the addition, subtraction, multiplication and divide operation. One or more operation can be applied at a time.

If f and g are two functions they can be written as,

\(\left(f+g\right)\left(x\right),\left(f-g\right)\left(x\right),\left(f\right)\left(x\right),\left({f}/{g}\; \right)\left(x\right)\)

Given

\(\begin{array}{l}g(x) = 2x + 1 \\h(x) = 4{x^2} + 4x + 7\end{array}\)

\(\begin{array}{l}f(x) = 3x + 5 \\h(x) = 3{x^2} + 3x + 2\end{array}\)

(a)Step 3: Function f

\(\begin{array}{l}g(x) = 2x + 1 \\h(x) = 4{x^2} + 4x + 7\end{array}\)

By seeing the variable terms in g and h, it is clear that we have to square g to get the terms of h. Now,

\(f(x) = {x^2} + c\)

Then,

\(\begin{array}{l}\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\\\left( {f \circ g} \right)\left( x \right) = f\left( {2x + 1} \right)\\\left( {f \circ g} \right)\left( x \right) = {\left( {2x + 1} \right)^2} + c\\\left( {f \circ g} \right)\left( x \right) = 4{x^2} + 4x + \left( {1 + c} \right)\end{array}\)

Now,

\(\begin{aligned}1 + c = 7\\c = 6\\f(x) = {x^2} + 6\end{aligned}\)

(b)Step 4: Function g

We need to find a function g like,

\(\begin{array}{l}f\left( {g\left( x \right)} \right) = 3\left( {g\left( x \right)} \right) + 5\\f\left( {g\left( x \right)} \right) = h\left( x \right)\end{array}\)

But,

\(\)

\(\begin{array}{l}h(x) = 3{x^2} + 3x + 2\\h(x) = 3\left( {{x^2} + x} \right) + 2\\h(x) = 3\left( {{x^2} + x - 1} \right) + 5\\g(x) = \left( {{x^2} + x - 1} \right)\end{array}\) \(\)