Graph $f ( x ) = \left\{ \begin{array} { l c } 4 x + 3 & x

Question

Graph $f ( x ) = \left\{ \begin{array} { l c } 4 x + 3 & x < 0 \ x + 3 & x \geq 0 \end{array} \right.$ and find the limit of $f ( x )$ as $x$ approaches $0 .$

A) 15 
B) 3 
C) 12 
D) $- 3$ 
E) limit does not exist

Quizplus · Accepted Answer

For values of $x$ less than 0, the function is a line with slope 4 passing through the point (0,3). For values of $x$ greater than or equal to 0, the function is a line with slope 1 passing through the point (0,3). Therefore, the limit of $f(x)$ as $x$ approaches 0 from both the left and the right is just $f(0)=3$.

Graph f(x)={4x+3x<0x+3x≥0f ( x ) = \left\{ \begin{array} { l c } 4 x + 3 & x < 0 \\ x + 3 & x \geq 0 \end{array} \right.f(x)={4x+3x+3​x<0x≥0​

Graph $f ( x ) = \left\{ \begin{array} { l c } 4 x + 3 & x < 0 \\ x + 3 & x \geq 0 \end{array} \right.$