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F'(x) = -1/x^5, f(-1) = -7/4
JMAP math differential-equation
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f'(x) = -1/x5, f(-1) = -7/4

A) f(x) = -1/x2 - 2, x < 0

B) f(x) = -3/x, x > 0

C) f(x) = 1/4x4 - 3, x > 0

D) f(x) = 1/4x4 - 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/x5

The equation above can be written as:

f'(x) = -x-5

Integrate both sides with respect to x

∫f'(x) = ∫-x-5

∫f'(x) = -∫x-5

f(x) = -(x-5 + 1) / -5+1 + c

where c is the integration constant.

f(x) = -x-4/-4 + c

f(x) = x-4/4 + c [The negative sign in the numerator and denominator cancels each other out].

f(x) = 1/4x4 + c ----------- eqn(1)

At f(-1) = -7/4 [that is, substitute x = -1 and equate to -7/4]

eqn(1) becomes,

-7/4 = 1/4(-1)4 + c

-7/4 = 1/4 + c

subtract 1/4 from both sides.

c = -7/4 - 1/4

c = -2

Thus, eqn(1) becomes:

f(x) = 1/4x4 - 2, x < 0

Ans D

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