The cost of purifying a gallon of water to a purity of x percent is shown below.C(x) = 100100 − x  cents for 50 ≤ x < 100(a) Find the instantaneous rate of change of the cost with respect to purity.C'(x) = −100(100−x2)​+1100−x​   (b) Evaluate this rate of change for a purity of 95%.C'(95) =  centsInterpret your answer.The cost is  ---Select--- increasing decreasing by this amount per additional percent of purity.(c) Evaluate this rate of change for a purity of 98%.C'(98) =  centsInterpret your answer.The cost is  ---Select--- increasing decreasing by this amount per additional percent of purity.

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Asked Oct 3, 2019
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The cost of purifying a gallon of water to a purity of x percent is shown below.

C(x) = 
100
100 − x
 
 cents for 
50 ≤ x < 100
(a) Find the instantaneous rate of change of the cost with respect to purity.
C'(x) = 
−100(100−x2)​+1100−x​
 
 
 


(b) Evaluate this rate of change for a purity of 95%.
C'(95)
 =  cents

Interpret your answer.
The cost is  ---Select--- increasing decreasing by this amount per additional percent of purity.

(c) Evaluate this rate of change for a purity of 98%.
C'(98)
 =  cents

Interpret your answer.
The cost is  ---Select--- increasing decreasing by this amount per additional percent of purity.
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100 50sx<100 C(x) -. 100 x C'(x)=(100-x)-(-1)(100) (100 x) 2 100 C'x)=100-x)^2

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100 C'(x)= 100-x)^ 2 100 = 4 25 100 C95)-100-95)^

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