Two piece wise functions

Algebra Level 4

If a step function, S ( x ) = { 0 for x ≤ 0 1 for x > 0 S(x)= \begin{cases} 0 &\text{for} & x \leq 0 \\ 1 &\text{for} & x > 0 \end{cases}
Then a piece wise function f ( x ) = { x 2 for 1 < x ≤ 2 4 x for x > 2 f(x)=\begin{cases} x^2 & \text{for} & 1<x \leq 2 \\ 4x & \text{for} & x > 2 \end{cases} can be written in terms of S as f ( x ) = x 2 + ( a x − x 2 ) S ( x + b ) f(x)=x^2+(ax-x^2)S(x+b)
Then, find the value of a+b


The answer is 2.

Shanthanu Rai by Shanthanu Rai , 21, India

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1 solution

展豪 張
Apr 10, 2016

We can 'construct' f ( x ) f(x) step by step:
S ( x ) = { 0 for x ≤ 0 1 for x > 0 S(x)=\begin{cases}0&\text{for}&x\leq 0\\1&\text{for}&x\gt 0\end{cases}
S ( x − 2 ) = { 0 for x ≤ 2 1 for x > 2 S(x-2)=\begin{cases}0&\text{for}&x\leq 2\\1&\text{for}&x\gt 2\end{cases}
x 2 + S ( x − 2 ) = { x 2 for x ≤ 2 x 2 + 1 for x > 2 x^2+S(x-2)=\begin{cases}x^2&\text{for}&x\leq 2\\x^2+1&\text{for}&x\gt 2\end{cases}
x 2 + ( 4 x − x 2 ) S ( x − 2 ) = { x 2 for x ≤ 2 4 x for x > 2 x^2+(4x-x^2)S(x-2)=\begin{cases}x^2&\text{for}&x\leq 2\\4x&\text{for}&x\gt 2\end{cases}
Just 'chop' the x ≤ 1 x\leq 1 cases out and we are done:
x 2 + ( 4 x − x 2 ) S ( x − 2 ) = { x 2 for 1 < x ≤ 2 4 x for x > 2 x^2+(4x-x^2)S(x-2)=\begin{cases}x^2&\text{for}&1\lt x\leq 2\\4x&\text{for}&x\gt 2\end{cases}
∴ a = 4 , b = − 2 \therefore a=4,b=-2
a + b = 2 a+b=2


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