f'(x) = -2/x
Integrate both sides with respect to x.
∫f'(x) = ∫-2/x
∫f'(x) = -2∫1/x
f(x) = - 2lnx + c ----------- eqn(1)
where c is the integration constant.
At f(2) = -2ln2 [that is, substitute x = 2 and equate to -2ln2]
-2ln2 = -2ln(2) + c
Add 2ln2 to both sides,
-2ln2 + 2ln2 = - 2ln2 + 2ln2 + c
0 = 0 + c
c = 0
Thus, eqn(1) becomes:
f(x) = -2lnx + 0
f(x) = -2lnx, x > 0 [x > 0 since x = 2 and 2 is greate than 0]
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