Answer:
Required critical point is (5, -25) which takes local minima.
Step-by-step explanation:
Given function is,
[tex]f(x)=x^2-10x\hfill (1)[/tex]
[tex]f'(x)=0\implies 2x-10=0\implies x=5[/tex]
Now substitite this value in (1) we get,
[tex]y=f(x)=(5)^2-(10\times 5)=-25[/tex]
Therefore the critical point is (5,-25).
[tex]f''(5)=2>0[/tex]
since double derivative at x=5 is geater than zero so critical point (5,-25) takes local minima.