Question:

Step 1: The function $f(x)$ is defined as $$f(x) = \int_0^x tg'(t)\,dt \quad \text{for } x \geq 0.$$ Step 2: Apply integration by parts to $f(x)$. Let $u=t$ and $dv=g'(t)\,dt$, so $du=dt$ and $v=g(t)$. $$f(x) = \left[tg(t)\right]_0^x - \int_0^x g(t)\,dt = xg(x) - 0 \cdot g(0) - \int_0^x g(t)\,dt = xg(x) - \int_0^x g(t)\,dt.$$ Step 3: Differentiate $f(x)$ with respect to $x$. Using the product rule for $xg(x)$ and the Fundamental Theorem of Calculus for the integral term: $$f'(x) = \frac{d}{dx}\left(xg(x) - \int_0^x g(t)\,dt\right) = \left(1 \cdot g(x) + xg'(x)\right) - g(x) = xg'(x).$$ Since $g'(x)$ exists for all $x \geq 0$ (as $g'(x) < 0$ is given), $f'(x)$ exists for all $x > 0$. Step 4: Since $f'(x)$ exists for all $x > 0$, $f(x)$ is differentiable for all $x > 0$. A differentiable function is continuous, so $f(x)$ is continuous for all $x > 0$. Step 5: Analyze the monotonicity of $f(x)$. From Step 3, $f'(x) = xg'(x)$. For $x > 0$, it is given that $g'(x) < 0$. Therefore, for $x > 0$, $f'(x) = xg'(x) < 0$. A function with a negative derivative on an interval is strictly decreasing on that interval. Thus, $f(x)$ is strictly decreasing for $x > 0$, which implies $f(x)$ is not increasing.