Funky Distance Function

The function goes smoothly from limit in minus infinity of $\frac12$ to a zero at $x=0$ . It remain zero until $x=1$ . It rises with slope 1 to the value of 1 at $x=2$ . It remains at 1 until $x=3$ . From the point $(3, 1)$ , it drops smoothly to its limit of $\frac12$ at infinity.

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Statement A:

The range of the function is $[0, 1]$ and $f(x)=0$ on $[0, 1]$ , so statement A is true.

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Statement B:

The function is symmetrical around the point $(1.5, 0.5)$ . This is easy to see for values from $0$ to $3$ .

Above $x=3$ the function $f(x)=\frac{x-1}{x-1+x-3}=\frac{x-1}{2x-4}$ .

Below zero $f(x)=\frac {-x}{-x-x+2}=\frac {-x}{-2x+2}$ . This can be flipped around $y$ -axis to make $\frac x{2x+2}$ , then around $x$ -axis to get $\frac {-x}{2x+2}$ , add 1 to that $\frac {-x}{2x+2}+1=\frac {-x}{2x+2}+\frac {2x+2}{2x+2}=\frac {x+2}{2x+2}$ , and finally slide that 3 units to the right (to the location of the part of the function we are trying to compare this to) $\frac {x-3+2}{2(x-3)+2}=\frac {x-1}{2x-4}$ . This is indeed the value of $f(x)$ in that region, so the symmetry holds.

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Statement C:

The function is equal zero on the interval $[0, 1]$ . The first derivative is zero on $(0, 1)$ and $(2, 3)$ . The second derivative is zero on $(0, 1), (1, 2)$ , and $(2, 3)$ . So the statement is true with $a=1$ .