determine all inflection points of f(x)=(18x + 9)(x - 5)^(7/3)+3x - 1. as your answer, please enter the sum…

Explanation:

Step1: Find the first - derivative

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime) where (u = 18x + 9), (u^\prime=18), (v=(x - 5)^{7/3}), (v^\prime=\frac{7}{3}(x - 5)^{4/3}). The derivative of (3x-1) is 3. (f^\prime(x)=18(x - 5)^{7/3}+(18x + 9)\times\frac{7}{3}(x - 5)^{4/3}+3) (=18(x - 5)^{7/3}+(42x + 21)(x - 5)^{4/3}+3) (=(x - 5)^{4/3}[18(x - 5)+42x + 21]+3) (=(x - 5)^{4/3}(18x-90 + 42x + 21)+3) (=(x - 5)^{4/3}(60x-69)+3)

Step2: Find the second - derivative

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime) again, where (u = 60x-69), (u^\prime = 60), (v=(x - 5)^{4/3}), (v^\prime=\frac{4}{3}(x - 5)^{1/3}). (f^{\prime\prime}(x)=60(x - 5)^{4/3}+(60x - 69)\times\frac{4}{3}(x - 5)^{1/3}) (=(x - 5)^{1/3}[60(x - 5)+\frac{4}{3}(60x - 69)]) (=(x - 5)^{1/3}(60x-300 + 80x - 92)) (=(x - 5)^{1/3}(140x-392))

Step3: Set the second - derivative equal to zero

((x - 5)^{1/3}(140x - 392)=0) We have two cases: Case 1: ((x - 5)^{1/3}=0), then (x = 5). Case 2: (140x-392 = 0), then (140x=392), (x=\frac{392}{140}=\frac{14}{5}=2.8)

Answer:

(5 + 2.8=7.8)