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It is given that,f(x)=sqrt(16-x^(2))" on "[0,4]The mean value theorem states, If f(x) be a function suchult,(i) f(x) is continuous in the closed interval [a,b] and(ii) f^(')(x) exists in the open interval (a,b), thow there exlsts at least one point c in (a,b) such that,f(b)-f(a)=(b-a)f^(')(c)Now,f(x)=sqrt(16-x^(2))(i) The function defined, if16-x^(2) >= 0" or "-4 <= x <= 4since the given interval [0,4] is the suoset of -4 <= x <= 4 hence the given function is continuous in closed intorval [0,4](ii) Take, f(x)=sqrt(16-x^(2))now, differentiating with res ... See the full answer