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Determine algebraically if f(x)=√1-x is a function even, odd, or neither.

Sagot :

we have the function

[tex]f(x)=\sqrt[]{1-x}[/tex]

step 1

Verify if the function is even

Remember that

If a function satisfies f(−x) = f(x) for all x it is said to be an even function

so

[tex]\begin{gathered} f(-x)=\sqrt[]{1-(-x)} \\ f(-x)=\sqrt[]{1+x} \end{gathered}[/tex]

therefore

f(x) is not equal to f(-x)

the function is not even

step 2

Verify if the function is odd

Remember that

A function is odd if −f(x) = f(−x), for all x

[tex]\begin{gathered} -f(x)=-\sqrt[]{1-x} \\ f(-x)=\sqrt[]{1+x} \end{gathered}[/tex]

therefore

-f(x) is not equal to f(-x)

The function is not odd

therefore

the answer is neither