Question:

Step 1: Differentiate both sides of the given integral equation with respect to $x$. $$ \frac{d}{dx}\int_0^x t f(x-t)\, dt = \frac{d}{dx}(e^{2x} - 2x - 1) $$ $$ \frac{d}{dx}\int_0^x t f(x-t)\, dt = 2e^{2x} - 2 $$ Step 2: Rewrite the integral using the substitution $u = x - t$. Then $t = x - u$ and $dt = -du$. When $t=0$, $u=x$. When $t=x$, $u=0$. $$ \int_0^x t f(x-t)\, dt = \int_x^0 (x-u) f(u)\, (-du) = \int_0^x (x-u) f(u)\, du $$ $$ \int_0^x (x-u) f(u)\, du = x\int_0^x f(u)\,du - \int_0^x u f(u)\,du $$ Step 3: Differentiate the rewritten form of the integral with respect to $x$. $$ \frac{d}{dx}\left[x\int_0^x f(u)\,du - \int_0^x u f(u)\,du\right] $$ Applying the product rule and the Fundamental Theorem of Calculus: $$ \left(\int_0^x f(u)\,du + x f(x)\right) - x f(x) = \int_0^x f(u)\,du $$ Equating this result with the differentiation from Step 1: $$ \int_0^x f(u)\,du = 2e^{2x} - 2 $$ Step 4: Differentiate the result from Step 3 with respect to $x$ to find $f(x)$. $$ f(x) = \frac{d}{dx}(2e^{2x} - 2) = 4e^{2x} $$ Step 5: Evaluate $f(0)$. $$ f(0) = 4e^{2(0)} = 4e^0 = 4 $$ Step 6: Evaluate $\lim_{x \to 0} \dfrac{f(x)-4}{x}$. $$ \lim_{x \to 0} \frac{4e^{2x}-4}{x} = \lim_{x \to 0} \frac{4(e^{2x}-1)}{x} $$ Using the standard limit $\lim_{y \to 0} \frac{e^y-1}{y} = 1$: $$ \lim_{x \to 0} \frac{4(e^{2x}-1)}{x} = 4 \cdot \lim_{x \to 0} \frac{e^{2x}-1}{x} = 4 \cdot \lim_{x \to 0} \frac{2(e^{2x}-1)}{2x} = 4 \cdot 2 \cdot 1 = 8 $$ Step 7: Find $(f^{-1})'(4)$. First, find $x_0$ such that $f(x_0) = 4$. $$ 4e^{2x_0} = 4 \implies e^{2x_0} = 1 \implies 2x_0 = 0 \implies x_0 = 0 $$ Next, find $f'(x)$. $$ f'(x) = \frac{d}{dx}(4e^{2x}) = 8e^{2x} $$ Evaluate $f'(x_0) = f'(0)$. $$ f'(0) = 8e^{2(0)} = 8e^0 = 8 $$ By the inverse function theorem, $(f^{-1})'(4) = \frac{1}{f'(0)}$. $$ (f^{-1})'(4) = \frac{1}{8} $$ Step 8: Find the derivative of $f(x)$ with respect to $e^x$ at $x=0$. Let $y = e^x$. We want to find $\frac{df}{dy}$. Using the chain rule: $\frac{df}{dy} = \frac{df/dx}{dy/dx}$. We have $f'(x) = 8e^{2x}$ and $\frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x$. $$ \frac{df}{dy} = \frac{8e^{2x}}{e^x} = 8e^x $$ Evaluate this at $x=0$: $$ \left.\frac{df}{dy}\right|_{x=0} = 8e^0 = 8 $$