Given:
[tex]f(x)=\begin{cases}-\dfrac{1}{2}x^2 & 0<x\leq 3 \\ x^3+2x & x>3\end{cases}[/tex]
To find:
The value of f(3).
Step-by-step explanation:
For x=3, we have
[tex]f(x)=-\dfrac{1}{2}x^2[/tex]
Putting x=3 in the above function, we get
[tex]f(3)=-\dfrac{1}{2}(3)^2[/tex]
[tex]f(3)=-\dfrac{1}{2}(9)[/tex]
[tex]f(3)=-\dfrac{9}{2}[/tex]
Therefore, the value of f(3) is [tex]-\dfrac{9}{2}[/tex]. So, the correct option is a.
