Find f(x) if f'(x)= #(e^(2x) +4e^-x)/e^x# and f(ln2)=2. Do u just find the anti-derivative for numerator and denominator separately?

Question

Christopher Agresta · Accepted Answer

No you can't do it that way.

The derivative of a quotient is not just the quotient of the derivatives, so, likewise, the anti derivative of a quotient is not just the quotient of antiderivatives.. In general, finding antiderivatives is much harder than finding derivatives. (In fact some antiderivatives cannot be nicely expressed.) Noe way to find the antiderivative of f'(x)= #(e^(2x) +4e^-x)/e^x# is to rewrite it. f'(x)= #(e^(2x) +4e^-x)/e^x = e^(2x)/e^x + (4e^(-x))/e^x# # = e^x + 4e^(-2x)# Now, the antiderivative of a sum is just the sum of the antiderivative (just as the derivative of a sum is he sum of the derivatives). So, #f(x) = e^x -2e^(-2x) +C# (Check the answer by differentiating.) Now use the given point to find #C#.

William Abdalla · Answer

undefined To find $ f(x) $, integrate $ f'(x) $ with respect to $ x $, ensuring to apply the initial condition $ f(\ln 2) = 2 $. You can separately integrate the numerator and denominator and then divide the results. Therefore,

$$ f(x) = \frac{{\int (e^{2x} + 4e^{-x}) \, dx}}{{\int e^x \, dx}} $$

Integrating the numerator:

$$ \int (e^{2x} + 4e^{-x}) \, dx = \frac{1}{2} e^{2x} - 4e^{-x} + C_1 $$

Integrating the denominator:

$$ \int e^x \, dx = e^x + C_2 $$

Dividing the results:

$$ f(x) = \frac{\frac{1}{2} e^{2x} - 4e^{-x} + C_1}{e^x + C_2} $$

Using the initial condition $ f(\ln 2) = 2 $:

$$ 2 = \frac{\frac{1}{2} e^{2\ln 2} - 4e^{-\ln 2} + C_1}{e^{\ln 2} + C_2} $$

$$ 2 = \frac{\frac{1}{2} \cdot 2^2 - 4 \cdot 2^{-1} + C_1}{2 + C_2} $$

$$ 2 = \frac{2 - 2 + C_1}{2 + C_2} $$

$$ 2 = \frac{C_1}{2 + C_2} $$

$$ 4 + 2C_2 = C_1 $$

Substituting back into $ f(x) $:

$$ f(x) = \frac{\frac{1}{2} e^{2x} - 4e^{-x} + (4 + 2C_2)}{e^x + C_2} $$

$$ f(x) = \frac{1}{2} e^{x} - 4e^{-x} + 2 + C_2 $$

Therefore,

$$ f(x) = \frac{1}{2} e^{x} - 4e^{-x} + 2 + C_2 $$