Science:Math Exam Resources/Courses/MATH110/December 2014/Question 03 (a)/Solution 1

For $x\leq 0$, $f(x)=e^{x}$ has horizontal asymptote $y=0$ because of $\lim _{x\to -\infty }e^{x}=0$, is increasing and passes though $f(0)=e^{0}=1$.

On the other hand, for $0<x<4$, the graph of $f(x)={\sqrt {x}}$ is a piece of parabola which satisfies $\lim _{x\to 0^{+}}f(x)=\lim _{x\to 0^{+}}{\sqrt {x}}=0$, $f(1)=1$, and $\lim _{x\to 4^{-}}f(x)=\lim _{x\to 4^{-}}{\sqrt {x}}=2$.

Finally, for $x\geq 4$, the graph of $f(x)=2x-6$ is a straight line passing through $(4,f(4))=(4,2)$.

Based on this information, we can draw the graph of $f$ as follows;