how would you limit the domain to make this function one-to-one and still have the same range? f(x) = x⁴ x ≥ ?

Explanation:

Step1: Analyze the function ( f(x) = x^4 )

The function ( f(x) = x^4 ) is a quartic function. Its graph is symmetric about the y - axis (since ( f(-x)=(-x)^4=x^4 = f(x) )). The range of ( f(x)=x^4 ) is ( y\geq0 ) because any real number raised to the fourth power is non - negative.

Step2: Recall the definition of a one - to - one function

A function is one - to - one (injective) if for every ( y ) in the range, there is exactly one ( x ) in the domain such that ( f(x)=y ). For the function ( f(x)=x^4 ), to make it one - to - one, we need to restrict the domain so that the function is either strictly increasing or strictly decreasing on that domain.

Step3: Analyze the monotonicity of ( f(x)=x^4 )

We find the derivative of ( f(x) ) using the power rule. The derivative ( f^\prime(x)=4x^3 ).

• When ( x>0 ), ( f^\prime(x)=4x^3>0 ), so the function is strictly increasing on the interval ( (0,+\infty) ).
• When ( x < 0 ), ( f^\prime(x)=4x^3<0 ), so the function is strictly decreasing on the interval ( (-\infty,0) ).

To keep the same range (( y\geq0 )) and make the function one - to - one, we can restrict the domain to ( x\geq0 ) (the interval where the function is strictly increasing). If we choose ( x\geq0 ), for each non - negative ( y ) in the range, there is exactly one non - negative ( x ) (since if ( y = x^4 ), then ( x=\sqrt[4]{y} ) when ( x\geq0 )) such that ( f(x)=y ). If we chose ( x\leq0 ), the range would still be ( y\geq0 ), but the function would be strictly decreasing, and ( x=\sqrt[4]{y} ) would not be in the domain ( x\leq0 ) (since ( \sqrt[4]{y}\geq0 ) for ( y\geq0 )). So the value to fill in the box is 0.

Answer:

( 0 )