1. **Stating the problem:** We are given a function $f$ with conditions involving $f^{-1}$, $f(x) + 3x > 0$, $f(3) = -8$, and another function $h(x) = (x^2 + 1) \cdot \ln(f(x) + 3x)$ with $h'(3) = 25$. We want to analyze these and find relevant derivatives or values. 2. **Given information:** - $f(3) = -8$ - $f(x) + 3x > 0$ (so the argument of the logarithm in $h(x)$ is positive) - $h(x) = (x^2 + 1) \cdot \ln(f(x) + 3x)$ - $h'(3) = 25$ 3. **Find $h'(x)$ using the product and chain rules:** $$h'(x) = \frac{d}{dx}[(x^2 + 1) \cdot \ln(f(x) + 3x)] = (2x) \cdot \ln(f(x) + 3x) + (x^2 + 1) \cdot \frac{1}{f(x) + 3x} \cdot (f'(x) + 3)$$ 4. **Evaluate at $x=3$:** - Compute $f(3) + 3 \cdot 3 = -8 + 9 = 1$ - Compute $\ln(1) = 0$ So, $$h'(3) = 2 \cdot 3 \cdot 0 + (3^2 + 1) \cdot \frac{1}{1} \cdot (f'(3) + 3) = 10 \cdot (f'(3) + 3)$$ 5. **Use the given $h'(3) = 25$ to solve for $f'(3)$:** $$25 = 10 \cdot (f'(3) + 3) \implies f'(3) + 3 = \frac{25}{10} = 2.5$$ $$f'(3) = 2.5 - 3 = -0.5$$ **Final answer:** $$f'(3) = -0.5$$