SOLUTION
(7). For a function to be even, then
[tex]f(x)=f(-x)[/tex]
And if a function is odd, then
[tex]f(x)=-f(-x)[/tex]
Now, let us check
[tex]\begin{gathered} f(x)=x\sqrt[]{4-x^2} \\ f(-x)\text{ becomes , that is replacing }x\text{ with -}x \\ f(-x)=-x\sqrt[]{4-(-x)^2} \\ f(-x)=-x\sqrt[]{4-x^2} \\ So,\text{ }f(x)=x\sqrt[]{4-x^2}\text{ and }f(-x)=-x\sqrt[]{4-x^2}\text{ } \\ we\text{ can s}ee\text{ that }f(x)\ne f(-x) \end{gathered}[/tex]
Hence, the function is not even.
Let's check for odd
[tex]\begin{gathered} \text{For odd} \\ f(x)=-f(-x) \\ f(x)=x\sqrt[]{4-x^2}\text{ } \\ -f(-x)=-(-x\sqrt[]{4-(-x)^2} \\ -f(-x)=-(-x\sqrt[]{4-x^2} \\ -f(-x)=x\sqrt[]{4-x^2} \\ \text{Now we can s}ee\text{ that }f(x)=x\sqrt[]{4-x^2}\text{ and }-f(-x)=x\sqrt[]{4-x^2} \\ So,\text{ }f(x)=-f(-x) \end{gathered}[/tex]
Hence the function is odd
