Given f(x) = int_{-2}^{x} t cdot g'(t) , dt f | vedclass

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Question

(C) Using the Fundamental Theorem of Calculus,we differentiate $f(x)$ with respect to $x$:$f'(x) = \frac{d}{dx} \int_{-2}^{x} t \cdot g'(t) \, dt = x

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$f(x)$ is increasing for $x > 0$ and decreasing for $x \in [-2, 0)$.

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(C) Using the Fundamental Theorem of Calculus,we differentiate $f(x)$ with respect to $x$:$f'(x) = \frac{d}{dx} \int_{-2}^{x} t \cdot g'(t) \, dt = x \cdot g'(x)$Since $g$ is an increasing function,$g'(x) \geq 0$ for all $x \geq -2$.Now,analyze the sign of $f'(x) = x \cdot g'(x)$:$1$. For $x \in [-2, 0)$,$x $2$. For $x > 0$,$x > 0$ and $g'(x) \geq 0$,so $f'(x) \geq 0$. Thus,$f(x)$ is an increasing function on $(0, \infty)$.Therefore,$f(x)$ is decreasing for $x \in [-2, 0)$ and increasing for $x > 0$.

Given $f(x) = \int_{-2}^{x} t \cdot g'(t) \, dt$ for $x \geq -2$,where $g$ is an increasing function,then:

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