A. Domain is 0≤x≤100 because x is a percentage. However, if c(x)≥0 then there must be a minimum value for x. This implies there is no cost for removing less than the minimum percentage of pollutants.
If c(x)=50x^2-100x-4900, and c(x)=0, x^2-2x-98=0, x^2-2x+1=99; (x-1)^2=99 and minimum x=1+√99=1+3√11=10.95 approx, making the domain 1+3√11≤x≤100 so that 0≤c(x)≤485100 (pesos). For removing less than 10.95% of pollutants there is no cost.
B. If c(x)=50x^2-100x-4900, then putting x=50, 70, 90 and 99.5 into the function we get:
c(50)=115100 pesos; c(70)=233100 pesos; c(90)=391100 pesos; c(99.5)=480162.5 pesos.
c(x) would be negative if x=0 which suggests perhaps the function is meant to read differently or have the domain specified in A.
If the formula for c(x) has not been correctly interpreted, please resubmit your question with a clearer representation of c(x). However, I hope the method shown is helpful to you.