0

f'(x) = 1/x^{2}, f(-1) = -1

A) f(x) = -1/2x^{4} + 2, x < 0

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

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

D) f(x) = -1/x^{2} + 2, 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 • 26 June 2020

University of Benin • 26 June 2020

0

f'(x) = 1/x^{2}

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/x^{1} + 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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