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F'(x) = 1/x^2, f(-1) = -1
JMAP math differential-equation
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f'(x) = 1/x2, f(-1) = -1

A) f(x) = -1/2x4 + 2, x < 0

B) f(x) = -1/x - 2, x < 0

C) f(x) = 1/x - 2, x < 0

D) f(x) = -1/x2 + 2, x < 0

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peter
Asked by peterrep:202
School not set 26 June 2020

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judziy
Judziy answeredrep:213
University of Benin 26 June 2020
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f'(x) = 1/x2

The equation above can be written as:

f'(x) = x-2

Integrate both sides with respect to x

∫f'(x) = ∫x-2

∫f'(x) = ∫x-2

f(x) = (x-2 + 1) / -2+1 + c

where c is the integration constant.

f(x) = x-1/-1 + c

f(x) = -x-1/1 + c

f(x) = -1/x1 + c

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

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

eqn(1) becomes,

-1 = -1/(-1) + c

-1 = 1 + c

Subtract 1 from both sides.

c = -1 - 1

c = -2

Thus, eqn(1) becomes:

f(x) = -1/x - 2, x < 0

Ans B

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