1. **Problem statement:** Given the function $g(x) = (x + 1)e^{ax} + b$ with $a,b \neq 0$, and the curve $(G)$ of $g$ has an inflection point at $x=1$ and a horizontal asymptote $y = -1$ as $x \to +\infty$. Show that $a = b = -1$. 2. **Formula and rules:** - Inflection point occurs where the second derivative $g''(x)$ changes sign, so $g''(1) = 0$. - Horizontal asymptote $y = -1$ at $+\infty$ means $\lim_{x \to +\infty} g(x) = -1$. 3. **Calculate limits and derivatives:** - $\lim_{x \to +\infty} g(x) = \lim_{x \to +\infty} (x+1)e^{ax} + b$. - Since $a \neq 0$, for the limit to be finite, $a$ must be negative so that $e^{ax} \to 0$ as $x \to +\infty$. - Then $\lim_{x \to +\infty} g(x) = 0 + b = b$. - Given the asymptote $y = -1$, so $b = -1$. 4. **Derivatives:** - First derivative: $$g'(x) = \frac{d}{dx}[(x+1)e^{ax} + b] = e^{ax} + (x+1)ae^{ax} + 0 = e^{ax}[1 + a(x+1)]$$ - Second derivative: $$g''(x) = \frac{d}{dx} \left(e^{ax}[1 + a(x+1)]\right) = e^{ax} \left(a[1 + a(x+1)] + a \right) = e^{ax} a [2 + a(x+1)]$$ 5. **Inflection point condition:** - At $x=1$, $g''(1) = 0$. - Since $e^{a \cdot 1} \neq 0$, we have: $$a[2 + a(1+1)] = 0 \implies a[2 + 2a] = 0$$ - Since $a \neq 0$, then: $$2 + 2a = 0 \implies 2a = -2 \implies a = -1$$ 6. **Conclusion:** - From step 3, $b = -1$. - From step 5, $a = -1$. **Final answer:** $a = b = -1$.