3. Write equations for two functions of even symmetry with 3 x-intercepts, two of which are x=1 and x=-1 (both multiplicity 1 ).
So, the two functions of even symmetry with 3 x-intercepts at x=1, x=-1 and x=0 are $\boxed{f(x) = x(x-1)(x+1)}$ and $\boxed{f(x) = 2x(x-1)(x+1)}$.
Step 1 :An even function is symmetric about the y-axis, which means that $f(x) = f(-x)$.
Step 2 :Given that the function has x-intercepts at x=1 and x=-1, we know that the function must be 0 at these points. Therefore, the function must have factors of $(x-1)$ and $(x+1)$.
Step 3 :Since the function is even, it must also have a x-intercept at x=0 to maintain symmetry about the y-axis. Therefore, the function must also have a factor of $x$.
Step 4 :Thus, one possible function is $f(x) = x(x-1)(x+1)$.
Step 5 :However, we need two functions. To get another function, we can simply multiply the first function by any non-zero constant. For example, if we multiply by 2, we get $f(x) = 2x(x-1)(x+1)$.
Step 6 :So, the two functions of even symmetry with 3 x-intercepts at x=1, x=-1 and x=0 are $\boxed{f(x) = x(x-1)(x+1)}$ and $\boxed{f(x) = 2x(x-1)(x+1)}$.