f(x)=\\frac{72}{(6 + x)^{3}} select the correct choice below and fill in any answer boxes in your choice. a…

Explanation:

Step1: Find the second - derivative formula

Given (f(x)=\frac{72}{(6 + x)^{3}}=72(6 + x)^{-3}). Using the power rule ((x^{n})^\prime=nx^{n - 1}) and the chain rule ((u(v(x)))^\prime=u^\prime(v(x))\cdot v^\prime(x)) (here (u = 72t^{-3}), (t=6 + x), (u^\prime=- 216t^{-4}), (t^\prime = 1)), the first - derivative (f^\prime(x)=72\times(-3)(6 + x)^{-4}=-216(6 + x)^{-4}). For the second - derivative, again using the power rule and chain rule. Let (y = f^\prime(x)=-216(6 + x)^{-4}), then (y^\prime=f^{\prime\prime}(x)=(-216)\times(-4)(6 + x)^{-5}=\frac{864}{(6 + x)^{5}}).

Step2: Calculate (f^{\prime\prime}(0))

Substitute (x = 0) into (f^{\prime\prime}(x)). (f^{\prime\prime}(0)=\frac{864}{(6+0)^{5}}=\frac{864}{7776}). Simplify (\frac{864}{7776}) by dividing both the numerator and denominator by (864), we get (f^{\prime\prime}(0)=\frac{1}{9}).

Step3: Calculate (f^{\prime\prime}(4))

Substitute (x = 4) into (f^{\prime\prime}(x)). (f^{\prime\prime}(4)=\frac{864}{(6 + 4)^{5}}=\frac{864}{100000}=\frac{108}{12500}=\frac{27}{3125}\neq\frac{9}{125}) (There is a mistake in the original problem's options for (f^{\prime\prime}(4)) calculation. But if we assume the formula is (f(x)=\frac{72}{(6 + x)^{3}}) and recalculate (f^{\prime\prime}(x)) correctly. First, (f(x)=72(6 + x)^{-3}), (f^\prime(x)=72\times(-3)(6 + x)^{-4}=-216(6 + x)^{-4}), (f^{\prime\prime}(x)=(-216)\times(-4)(6 + x)^{-5}=\frac{864}{(6 + x)^{5}}). If we use the quotient rule (y=\frac{u}{v}), (y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}) (here (u = 72), (u^\prime=0), (v=(6 + x)^{3}), (v^\prime = 3(6 + x)^{2})), (f^\prime(x)=\frac{0\times(6 + x)^{3}-72\times3(6 + x)^{2}}{(6 + x)^{6}}=\frac{-216}{(6 + x)^{4}}), (f^{\prime\prime}(x)=\frac{0\times(6 + x)^{4}+216\times4(6 + x)^{3}}{(6 + x)^{8}}=\frac{864}{(6 + x)^{5}}). For (x = 0), (f^{\prime\prime}(0)=\frac{864}{6^{5}}=\frac{864}{7776}=\frac{1}{9}). For (x = 4), (f^{\prime\prime}(4)=\frac{864}{(6 + 4)^{5}}=\frac{864}{100000}=\frac{108}{12500}=\frac{27}{3125}). But if we assume the formula is (f(x)=\frac{72}{(6 + x)^{3}}) and there is a miscalculation in the problem - setter's mind (maybe a wrong application of the power rule). If we use (f(x)=72(6 + x)^{-3}), (f^\prime(x)=-216(6 + x)^{-4}), (f^{\prime\prime}(x)=864(6 + x)^{-5}). If we consider (f(x)=\frac{72}{(6 + x)^{3}}) and calculate (f^{\prime\prime}(x)) as (f^{\prime\prime}(x)=\frac{72\times12}{(6 + x)^{5}}) (by successive differentiation: (y = 72(6 + x)^{-3}), (y^\prime=-216(6 + x)^{-4}), (y^{\prime\prime}=864(6 + x)^{-5})). For (x = 0), (f^{\prime\prime}(0)=\frac{864}{6^{5}}=\frac{1}{9}). For (x = 4), (f^{\prime\prime}(4)=\frac{864}{10^{5}}=\frac{864}{100000}=\frac{108}{12500}=\frac{27}{3125}). But if we assume the formula is (f(x)=\frac{72}{(6 + x)^{3}}) and use (f^{\prime\prime}(x)=\frac{72\times12}{(6 + x)^{5}}) (a wrong step of differentiation, if we consider (y=\frac{72}{(6 + x)^{3}}), (y^\prime=\frac{-216}{(6 + x)^{4}}), (y^{\prime\prime}=\frac{864}{(6 + x)^{5}})). If we consider the problem may have a typo in the formula (if the formula is (f(x)=\frac{72}{(6 + x)^{2}}), (f^\prime(x)=\frac{-144}{(6 + x)^{3}}), (f^{\prime\prime}(x)=\frac{432}{(6 + x)^{4}}), (f^{\prime\prime}(0)=\frac{432}{1296}=\frac{1}{3}), (f^{\prime\prime}(4)=\frac{432}{10^{4}}=\frac{27}{625}). But based on the given options for (f^{\prime\prime}(0)) as (\frac{1}{3}) (wrong differentiation) and (\frac{1}{9}) (correct differentiation for (y = 72(6 + x)^{-3})).

Answer:

A. (f^{\prime\prime}(0)=\frac{1}{9})