Given:
[tex]f(x)=\frac{1}{2}x^2-x-9[/tex]
Differentiate with respect to 'x'
[tex]\begin{gathered} f^{\prime}(x)=\frac{1}{2}(2x)-1 \\ f^{\prime}(x)=x-1 \end{gathered}[/tex]
Let f'(x)=0
[tex]\begin{gathered} x-1=0 \\ x=1 \end{gathered}[/tex][tex]\begin{gathered} f^{\doubleprime}(x)=1 \\ f^{\doubleprime}(x)>0 \end{gathered}[/tex]
Therefore, the function f(x) is minimum
[tex]\begin{gathered} f(1)=\frac{1}{2}(1)^2-1-9 \\ f(1)=\frac{1}{2}-1-9 \\ f(1)=\frac{1-2-18}{2} \\ f(1)=-\frac{19}{2} \\ f(1)=-9.5 \end{gathered}[/tex][tex]\text{Therefore, the minimum point is (1,-9.5)}[/tex]
Option d is the final answer.
