ISI2016-DCG-48

Question

The piecewise linear function for the following graph is

• A. $f(x)=\begin{cases} = x,x\leq-2 \\ =4,-2<x<3 \\  = x+1,x\geq 3\end{cases}$
• B. $f(x)=\begin{cases} = x-2,x\leq-2 \\ =4,-2<x<3 \\  = x-1,x\geq 3\end{cases}$
• C. $f(x)=\begin{cases} = 2x,x\leq-2 \\ =x,-2<x<3 \\  = x+1,x\geq 3\end{cases}$
• D. $f(x)=\begin{cases} = 2-x,x\leq-2 \\ =4,-2<x<3 \\  = x+1,x\geq 3\end{cases}$

Answer 1

For the interval   $x\leq 2$, the slope of the line is  $\frac{6-4}{-4-(-2)}=-1$

 

For the interval    $-2\lt x\lt 3$,   the slope of the line is   $0$,  and it is also clear that  the equation of the line in this case would be   $y=4$.

 

For the interval    $x\geq 3$,   the slope of the line  would be    $\frac{6-4}{5-3}=1$

 

We use point slope form of a line i.e.   $y-y_{1}=m\times (x-x_{1})$   for each of the interval given in the question, where  $m$  is the slope of the line,   $(x_{1},y_{1})$    is the point through which the line passes.

 

For the interval   $x\leq 2$,  the equation of the line is :     $y-4=-1\times (x-(-2))$    $\Rightarrow$     $y=2-x$

 

For the interval    $x\geq 3$,  the equation of the line is :   $y-4=1\times (x-3)$         $\Rightarrow$     $y=x+1$

 

Option D is the correct answer.

Comments

I think it would be better if you just check continuity of $f(x)$ at $x=-2,3$, because by looking at graph, we can say that $f(x)$ is continuous for every real $x$, and in every option just consists of polynomials and we all know that polynomials are continuous at every real x, so just try to check the continuity of f(x) at $x=-2,3$.