Christmas Streak 41/88: Func(ep)tion

Look no more, as we're only going to observe when $x=\pm 2.$

For $x=2,$

$\lim_{x\to2^{-}}f(x)=f(2) \\ 2+a=bf(1)+c \\ 2+a=b(a+1)+c$

For $x=-2,$

$\lim_{x\to -2^{+}}f(x)=f(-2) \\ -2+a=bf(-1)+c \\ -2+a=b(a-1)+c$

Subtracting the latter from the former, we see that $b=2$ and $a+c=0.$ $\Rightarrow~\dfrac{100}{a}+\dfrac{100}{c}=0$

Therefore the answer is $\boxed{50}.$

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Now to prove it's continuous.

for $2\le |x|<4,$ we notice that $f(x)=2f\left(\dfrac{x}{2}\right)+c=x+2a+c=x+a.$

Oh. This is the same as when $1\le |x|<2.$

We can immediately notice that this pattern will keep holding for all reals $x.$

Therefore the entire function is equal to $f(x)=x+a,$ proving $f(x)$ is continuous. $\square$

Any possible discontinuity would be due to discontinuity at $x = \pm 2$ . Therefore we require $a \pm 2 = b(a \pm 1) + c.$ Subtracting and adding the two equations gives $\begin{cases} 4 = 2b \\ 2a = 4a + 2c \end{cases}$ so that $b = 2$ and $c = -a$ . Therefore $\frac{100}a + \frac{100}b + \frac{100}c = \frac{100}a + 50 - \frac{100}a = \boxed{50}.$

Notice, both parts of the piecewise function are linear, therefore, they are themselves, continuous over their given domain. So for the piecewise function as a whole to be continuous, the 2 parts should cross at a common point, in other words, $x+a=bf(\frac{x}{2})+c \textrm{ at } x=\pm 2$ or $2+a=b(1+a)+c\textrm{ (1) and } a-2=b(a-1)+c \textrm{ (2)} \\ 4=2b \implies b=2 \textrm{ Subtract equation 2 from 1} \\ a+2=2+2a+c \\ a=2a+c \implies a=-c$ Putting this into the given expression, we get $\frac{100}{-c}+\frac{100}{2}+\frac{100}{c}=50+\frac{100}{c}-\frac{100}{c}=\boxed{50}$

$f(x)=\frac{bx}{2}+ab+c$

$f(2)=b+ab+c=2+a, f(-2)=-b+ab+c=-2+a$

Subtract two equations: $2b=4, b=2, a=-c$

Thus, since $\frac{100}{a}+\frac{100}{c}=0$ , the answer is $\frac{100}{b}=50$ .

We need to ensure that the graph is continuous at $x=-2$ and $x=2$ , which will occur when both pieces of the function are equal at the corresponding x-values.

$x=-2 \implies f(x)=-2+a=bf(-1)+c \implies -2+a=b(-1+a)+c \implies -2+a=-b+ab+c$

$x=2 \implies f(x)=2+a=bf(1)+c \implies 2+a=b(1+a)+c \implies 2+a=b+ab+c$

Subtracting the second equation from the first leaves us with $-4=-2b \implies b=2$ .

When we plug $2$ in for either equation, we will end up with $a+c=0$ , or $c=-a$ . This implies that $\frac{100}{a} + \frac{100}{c} = 0$ .

Therefore, our answer is $\frac{100}{2}=\boxed{50}$ .

The whole function is:

$f(x)=\cases{\begin{aligned}&x+a&&\text{for }|x|<2\\&2f\Big(\frac{x}{2}\Big)-a&&\text{for }|x|\ge2\end{aligned}}$

Now we must prove that this function is continuous. For some $|x|\ge2$ , the second expression will apply, and $f(\frac{x}{2})$ will be computed. This will happen again and again until we reach $f(\frac{x}{2^{n-1}})=2f(\frac{x}{2^n})-a$ and $|\frac{x}{2^n}| < 2$ . This will have an output of $2(\frac{x}{2^n}+a)-a = \frac{x}{2^{n-1}}+a$ . This will substitute $f(\frac{x}{2^{n-1}})$ in the previous expression, and we will be left with $\frac{x}{2^{n-2}}+a$ . The process continues recursively until we are left with $2(\frac{x}{2}+a)-a = x+a$ . Because this will be true for all real x-values, $f(x)$ will be a continuous line.