Functions: Overview - Who's Afraid of Functions?
If the symbol $$^*$$ represents a certain function for all real numbers, and $$(1^*)^* = 6$$, then which of the following could be the definition of $$^*$$ for all values of $$m$$?
Incorrect.
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Plug in $$m=1$$:
>$$(1^*) = 2(1)^3 = 2$$
>$$(1^*)^* = 2^* = 2(2)^3 = 16$$ and not 6.
__POE__ this answer choice and move on.
Incorrect.
[[snippet]]
Plug in $$m=1$$:
>$$(1^*) = (2\times 1)^3 = 2^3 = 8$$
>$$(1^*)^* = 8^* = (2\times 8)^3 = 16^3$$ and not 6.
__POE__ this answer choice and move on.
Incorrect.
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Plug in $$m=1$$:
>$$(1^*) = 3(1)^2 = 3$$
>$$(1^*)^* = 3^* = 3(3)^2 = 27$$ and not 6.
__POE__ this answer choice and move on.
Incorrect.
[[snippet]]
Plug in $$m=1$$:
>$$(1^*) = 1(1-1)(1+1) = 0$$
>$$(1^*)^* = 0^* = 0(0-1)(0+1)=0$$ and not 6.
__POE__ this answer choice and move on.
This answer is eliminated.
Correct.
[[snippet]]
Plug in $$m=1$$:
>$$(1^*) = 1(1+1) = 2$$.
Now that you have the value of $$1^*$$, use that value to find the value of $$(1^*)^*$$. Replace $$1^*$$ with 2 so that
>$$(1^*)^* = 2^* = 2(2+1) = 6$$.
Thus this function _could_ be true for all values of $$m$$.
$$m^* = 2m^3$$
$$m^* = m(m+1)$$
$$m^* = (2m)^3$$
$$m^* = 3m^2$$
$$m^* = m(m-1)(m+1)$$
