1. **State the problem:** Given the derivative $$f'(x) = 3e^{3x} + \frac{1}{4x - 2}$$, find an antiderivative $$f(x)$$ and the particular solution satisfying $$f\left(\frac{3}{4}\right) = 9.48774$$. 2. **Recall the formula:** To find an antiderivative, integrate the derivative: $$f(x) = \int f'(x)\,dx = \int \left(3e^{3x} + \frac{1}{4x - 2}\right) dx$$ 3. **Integrate each term separately:** - For $$3e^{3x}$$, use substitution: Let $$u = 3x$$, then $$du = 3 dx$$, so $$dx = \frac{du}{3}$$. $$\int 3e^{3x} dx = 3 \int e^{3x} dx = 3 \int e^u \frac{du}{3} = \int e^u du = e^u + C = e^{3x} + C$$ - For $$\frac{1}{4x - 2}$$, use substitution: Let $$v = 4x - 2$$, then $$dv = 4 dx$$, so $$dx = \frac{dv}{4}$$. $$\int \frac{1}{4x - 2} dx = \int \frac{1}{v} \frac{dv}{4} = \frac{1}{4} \int \frac{1}{v} dv = \frac{1}{4} \ln|v| + C = \frac{1}{4} \ln|4x - 2| + C$$ 4. **Combine the integrals:** $$f(x) = e^{3x} + \frac{1}{4} \ln|4x - 2| + C$$ 5. **Use the initial condition to find $$C$$:** Given $$f\left(\frac{3}{4}\right) = 9.48774$$, Calculate: $$e^{3 \times \frac{3}{4}} = e^{\frac{9}{4}}$$ $$\ln|4 \times \frac{3}{4} - 2| = \ln|3 - 2| = \ln 1 = 0$$ So, $$9.48774 = e^{\frac{9}{4}} + \frac{1}{4} \times 0 + C = e^{2.25} + C$$ Calculate $$e^{2.25} \approx 9.487735836$$ Then, $$C = 9.48774 - 9.487735836 = 0.000004164 \approx 0.00000416$$ 6. **Final answer:** $$f(x) = e^{3x} + \frac{1}{4} \ln|4x - 2| + 0.00000416$$