03.2functioneqns

# Functional Equations

Most equations are written in terms of unknown variables such as x.
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In this section, we look at equations that are written in terms of unknown functions such as f(x).
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The most common question will be to verify that a given f(x) satisfies the equation.
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Generally the technique is to substitute the function into the equation and simplify.

• If you end up with something which can never happen (like 2 = 3) then f(x) does not satisfy the equation.
• If you end up with something which is always true (like 5 = 5) then f(x) does satisfy the equation.
• If you end up with values for x (such as x = ±3) then f(x) only satisfies the equation for those values.

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#### Example 1

… … Given $f(x) = 2x + 3$,

… … determine if f(x) satisfies the equation $3f(x) = f(3x)$

Solution:

… … $3f(x) = 3(2x + 3)$

… … $f(3x) = 2(3x) + 3$

Equation is:

… … $3f(x) = f(3x)$

… … $3(2x + 3) = 2(3x) + 3$

… … $6x + 9 = 6x + 3$

… … $9 = 3$

… … This is never true so f(x) does not satisfy the equation

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#### Example 2

… … Given $f(x) = 5x$,

… … determine if f(x) satisfies the equation $3f(x) = f(3x)$

Solution:

… … $3f(x) = 3(5x)$

… … $f(3x) = 5(3x)$

Equation is:

… … $3f(x) = f(3x)$

… … $3(5x) = 5(3x)$

… … $15x = 15x$

… … This is true for all $x \in R$, so $f(x)$ does satisfy the equation

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## Functional Equations on the Classpad

The CAS calculator can verify functional equations.

First enter the function by typing:

… … Define $f(x) = 2x + 3$

… … Use the virtual keyboard ABC tab for the name of the function (f)

… … and use the variable keypad for x

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Then give the command to solve the equation by typing

… … Solve( $3f(x)=f(3x)$)

… … Solve is in virtual keyboard, Math1 tab, or the ACTION menu, EQUATION submenu

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… … the calculator responded with {No Solution}
… … so the function $f(x) = 2x+3$ does not satisfy the equation
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In a second example on the same screen,
… … I have defined the function $f(x) = 5x$.
… … The calculator has responded with ${x=x}$, which is always true,
… … so the function $f(x) = 5x$ does satisfy the equation.

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