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What is the inverse of the function 2x-e^2x?

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tufa | (Level 1) Honors

Posted March 19, 2011 at 12:12 AM via web

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What is the inverse of the function 2x-e^2x?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted March 19, 2011 at 12:22 AM (Answer #1)

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We'll put f(x) = 2x - e^2x.

According to the rule, f'(x)*[f^-1(x)]' = 1

[f^-1(x)]'= 1/f'(x)

f^-1(x) = Indefinite Integral of 1/f'(x)

We'll calculate f'(x) = (2x-e^2x)'

f'(x) = 2 - 2e^2x

[f^-1(x)]'= 1/(2 - 2e^2x)

We'll calculate the indefinite integral:

Int dx/(2 - 2e^2x) = (1/2)*Int dx/(1 - e^2x)

We'll put 1 - e^2x = t => e^2x = 1 - t

We'll differentiate:

2e^2x*dx = -dt

dx = -dt/2e^2x

dx = -dt/2(1 - t)

(1/2)*Int dx/(1 - e^2x) = (1/2)*Int -dt/2t*(1 - t)

(1/2)*Int -dt/2t*(1 - t) = (-1/4)*Int dt/t*(1 - t)

We'll decompose the fraction 1/t*(1-t) in a sum or difference of elementary fractions:

1/t*(1 - t) = A/t + B/(1-t)

1 = A - At + Bt

1 = t(B-A) + A

B-A = 0

A = B

A = 1 => B = 1

1/t*(1 - t) = 1/t + 1/(1-t)

(-1/4)*Int dt/t*(1 - t) = (-1/4)*[Int dt/t + Int dt/(1 - t)]

(-1/4)*Int dt/t*(1 - t) = (-1/4) [ln |t| + ln |1 - t|]  +C

(-1/4)*Int dt/t*(1 - t) = (-1/4) ln |t*(1-t)| + C

Int dx/(2 - 2e^2x)  = (-1/4) ln |(1 - e^2x)*(e^2x)| + C

The inverse function is f^-1(x) = -[ln |(1 - e^2x)*(e^2x)|]/4.

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