0

f'(x) = -1/x, f(3) = -ln(3) -1

A) f(x) = -3lnx - x, x < 0

B) f(x) = 2lnx + 1, x > 0

C) f(x) = -2lnx - x + 3, x < 0

D) f(x) = -lnx - 1, x > 0

Asked by judziy•rep:213

University of Benin • 26 June 2020

University of Benin • 26 June 2020

**Comments**

Peter answered•rep:202

School not set • 26 June 2020

School not set • 26 June 2020

0

f'(x) = -1/x

Integrate both sides with respect to x.

∫f'(x) = ∫-1/x

∫f'(x) = -∫1/x

f(x) = - lnx + c ----------- eqn(1)

where c is the integration constant.

At f(3) = -ln3 - 1 [that is, substitute x = 3 and equate to -ln3 - 1]

eqn(1) becomes,

-ln3 - 1 = -ln(3) + c

Add ln3 to both sides

-ln3 + ln3 - 1 = - ln3 + ln3 + c

0 - 1 = 0 + c

c = -1

Thus, eqn(1) becomes:

f(x) = -lnx - 1, x > 0 [x > 0 since x = 3 and 3 is greate than 0]

**Ans D**

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