Given the function
[tex]R(x)=50-0.1x^2[/tex]Then, R'(x) is:
[tex]R^{\prime}(x)=-2(0.1)x^{2-1}=-0.2x[/tex]Therefore:
A) R'(400)
[tex]R^{\prime}(400)=-0.2(400)=-80[/tex]Answer: - $80
B) R'(650)
[tex]R^{\prime}(650)=-0.2(650)=-130[/tex]Answer: -$130