I want to find a function $h(x)$ that always has exactly 1 distance to $f(x)$. This is called a parallel curve as I understand it, however, I don't want to do it this way, and I believe I solved the problem, however, my CAS software (Maple) cannot compute the function I get. I have done the following:
I have two values, $a$ and $b$, which represent a point $(a,b)$. I want to find the $b$-coordinate for any $a$-coordinate, i.e. a function $b(a)$. I start by assuming an $a$-value: $$a=1.5$$ I now define my function I want to stay at $1$ distance to $$f(x)=-x^2+2$$ I then define a function that can give me the distance between the point $(a, b)$ and the point $(x, f(x)$ $$g(x)=\sqrt{(a-x)^2+(b-f(x))^2}$$ I want to find an a-value for the point where the minimum of $g(x)$ is exactly 1, i.e. an $a$ for which $g(x)=1$ and $g'(x)=0$: $$g'(x)=0 \Leftrightarrow b=\frac{0.25(4x^3-6x-3)}{x}$$
I now plug this into $g(x)$ and set it to $1$: $$\sqrt{(1.5-x)^2+(- \frac{0.25(4x^3-6x-3}{x}-f(x))^2}=1 \Leftrightarrow x=0.69$$
I then plug this x-value into $g(x)$ and solve for $b$, giving me $b=2.11$
Now this is fine, but as mentioner earlier, I would like to find a function $b(a)$, so following the same steps: $$g'(x)=0 \Leftrightarrow b=\frac{-2x^3+a+3x}{2x}$$
Substituting:
$$\sqrt{((1.5)^2+(\frac{-2x^3+a+3x}{2x})^2}=1 \Leftrightarrow x=\text{Root of }(4z^4-12z^3+6z^2-2az+a^2)$$
And plugging this $x$ into the original function:
$$g(x)=1 \Leftrightarrow b=\text{Length of output exceeds limit of 1000000}$$
Since solving $g(x)=1$ for b depends on $a$, i must get a function $b(a)$, which is where Maple cant process anymore. Does this seem correct or am I missing something? Why can it not process?