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uysha [10]
4 years ago
14

Please show me how to do this.

Mathematics
1 answer:
Aloiza [94]4 years ago
7 0

Answer:

f(x) = \dfrac{1}{2}(x - \ln (x - 1) - 2)

Step-by-step explanation:

f'(x) = \dfrac{1}{2} - \dfrac{1}{2x - 2}

f(x) = \int (\dfrac{1}{2} - \dfrac{1}{2x - 2}) dx

f(x) = \int [\dfrac{1}{2} - \dfrac{1}{2(x - 2)}] dx

f(x) = \int \dfrac{1}{2}dx - \dfrac{1}{2} \int \dfrac{1}{x - 1} dx

f(x) = \dfrac{1}{2}x - \dfrac{1}{2} \ln (x - 1) + C

f(x) = \dfrac{x - \ln (x - 1)}{2} + C

f(2) = \dfrac{2 - \ln (2 - 1)}{2} + C = 0

f(2) = \dfrac{2 - \ln (1)}{2} + C = 0

f(2) = \dfrac{2}{2} + C = 0

f(2) = 1 + C = 0

C = -1

f(x) = \dfrac{x - \ln (x - 1)}{2} - 1

f(x) = \dfrac{1}{2}[x - \ln (x - 1) - 2]



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4 years ago
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