Dysfunctional with functions

[kitcat3200]

"True or false: For all real numbers a, f(h(a))=h(f(a)). If false, provide a counterexample."

I've reviewed definitions, tried to find/create examples and asked my dad (he took algebra 45 years ago!) but am still stumped. Any help or suggestions you can provide are much appreciated.

Many thanks!

 

[khavar]

I don't know the answer but let's try to figure this out.

Here is a possible counterexample:

f(a)=x^2+1
h(a)=x-1

f(h(a))=X^2-2x+2
h(f(a))=X^2

Plug in any real number. Let me know what you think, hopefully I'm not way off.

 

[pka]

"True or false: For all real numbers a, f(h(a))=h(f(a)). If false, provide a counterexample."

Let \(\displaystyle f(x)=x^2+1~\&~h(x)=(x+1)^2\) then look \(\displaystyle f(h(1))~\&~h(f(1))\).

 

[soroban]

Hello, kitcat3200!

Forty-five years ago, your father may not have learned functions.
But you must have learned about composite functions.

True or false: For all real numbers \(\displaystyle x,\;f\big(h(a)\big)\:=\:h\big(f(a)\big).\)
If false, provide a counterexample.

You should be able to come up with a counterexample.
Almost any two functions should work.

Let \(\displaystyle \begin{Bmatrix}f(x) \:=\:2x+1 \\ h(x) \:=\:3x-2\end{Bmatrix}\)

\(\displaystyle f\big(h(x)\big) \;=\;f(3x-2) \;=\;2(3x-2) + 1 \;=\;6x-4+1 \;=\;6x-3\)

\(\displaystyle h\big(f(x)\big) \;=\;h(2x+1) \;=\;3(2x+1)-2 \;=\;6x+3-2 \;=\;6x+1\)

They are not equal . . . The statement is false.