1. **State the problem:** We need to find real numbers $a$ and $b$ such that the piecewise function $$f(x) = \begin{cases} a x + \frac{b^{x-2}}{x+1} - 1, & x \leq 1 \\ b \sqrt{2x - 1} - a e^{x-1}, & x > 1 \end{cases}$$ is differentiable on $\mathbb{R}$. 2. **Key conditions for differentiability:** - The function must be continuous at $x=1$. - The left-hand derivative and right-hand derivative at $x=1$ must be equal. 3. **Continuity at $x=1$:** Evaluate both branches at $x=1$ and set equal: $$f(1^-) = a(1) + \frac{b^{1-2}}{1+1} - 1 = a + \frac{b^{-1}}{2} - 1 = a + \frac{1}{2b} - 1$$ $$f(1^+) = b \sqrt{2(1) - 1} - a e^{1-1} = b \sqrt{1} - a e^0 = b - a$$ Set equal for continuity: $$a + \frac{1}{2b} - 1 = b - a$$ 4. **Differentiability at $x=1$:** Calculate derivatives from left and right and set equal. - Left derivative $f'(x)$ for $x \leq 1$: $$f'(x) = a + \frac{d}{dx} \left( \frac{b^{x-2}}{x+1} \right)$$ Use quotient rule: Let $u = b^{x-2}$ and $v = x+1$. $$u' = b^{x-2} \ln b, \quad v' = 1$$ $$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{u'v - uv'}{v^2} = \frac{b^{x-2} \ln b (x+1) - b^{x-2} (1)}{(x+1)^2} = \frac{b^{x-2} ((x+1) \ln b - 1)}{(x+1)^2}$$ So, $$f'(x) = a + \frac{b^{x-2} ((x+1) \ln b - 1)}{(x+1)^2}$$ Evaluate at $x=1$: $$f'(1^-) = a + \frac{b^{-1} (2 \ln b - 1)}{4} = a + \frac{2 \ln b - 1}{4b}$$ - Right derivative $f'(x)$ for $x > 1$: $$f'(x) = b \frac{d}{dx} \sqrt{2x - 1} - a \frac{d}{dx} e^{x-1} = b \frac{1}{2 \sqrt{2x - 1}} (2) - a e^{x-1} = \frac{b}{\sqrt{2x - 1}} - a e^{x-1}$$ Evaluate at $x=1$: $$f'(1^+) = \frac{b}{\sqrt{1}} - a e^0 = b - a$$ 5. **Set derivatives equal for differentiability:** $$a + \frac{2 \ln b - 1}{4b} = b - a$$ 6. **Solve the system:** From continuity: $$a + \frac{1}{2b} - 1 = b - a \implies 2a + \frac{1}{2b} - 1 = b$$ From differentiability: $$a + \frac{2 \ln b - 1}{4b} = b - a \implies 2a + \frac{2 \ln b - 1}{4b} = b$$ Set right sides equal: $$2a + \frac{1}{2b} - 1 = 2a + \frac{2 \ln b - 1}{4b}$$ Simplify: $$\frac{1}{2b} - 1 = \frac{2 \ln b - 1}{4b}$$ Multiply both sides by $4b$: $$2 - 4b = 2 \ln b - 1$$ Rearranged: $$2 \ln b + 4b = 3$$ 7. **Solve for $b$ numerically:** Try $b=1$: $$2 \ln 1 + 4(1) = 0 + 4 = 4 \neq 3$$ Try $b=0.5$: $$2 \ln 0.5 + 4(0.5) = 2(-0.6931) + 2 = -1.386 + 2 = 0.614 < 3$$ Try $b=0.8$: $$2 \ln 0.8 + 4(0.8) = 2(-0.2231) + 3.2 = -0.446 + 3.2 = 2.754 < 3$$ Try $b=0.85$: $$2 \ln 0.85 + 4(0.85) = 2(-0.1625) + 3.4 = -0.325 + 3.4 = 3.075 > 3$$ So $b \approx 0.83$ (approximate root). 8. **Find $a$ using continuity equation:** $$2a + \frac{1}{2b} - 1 = b$$ $$2a = b + 1 - \frac{1}{2b}$$ $$a = \frac{b + 1 - \frac{1}{2b}}{2}$$ Substitute $b \approx 0.83$: $$a \approx \frac{0.83 + 1 - \frac{1}{2 \times 0.83}}{2} = \frac{1.83 - 0.602}{2} = \frac{1.228}{2} = 0.614$$ **Final answer:** $$a \approx 0.614, \quad b \approx 0.83$$