0

f'(x) = -2/x, f(2) = -2ln2

A) f(x) = -ln- x + 3, x < 0

B) f(x) = -2lnx + 2, x > 0

C) f(x) = -2lnx, x > 0

D) f(x) = -ln-x - 1, x < 0

Asked by peter•rep:202

School not set • 26 June 2020

School not set • 26 June 2020

**Comments**

Judziy answered•rep:213

University of Benin • 27 June 2020

University of Benin • 27 June 2020

0

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]

eqn(1) becomes,

-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]

**Ans C**

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