Let f(x) = int_{x^2}^{x^2+1} e^{-t^2} dt,for | vedclass

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(A) Given $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$. Applying the Leibniz rule for differentiation under the integral sign: $f'(x) = e^{-(x^2+1)^2} \cdo

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$(-\infty, 0]$

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(A) Given $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$. Applying the Leibniz rule for differentiation under the integral sign: $f'(x) = e^{-(x^2+1)^2} \cdot \frac{d}{dx}(x^2+1) - e^{-(x^2)^2} \cdot \frac{d}{dx}(x^2)$ $f'(x) = e^{-(x^2+1)^2} \cdot (2x) - e^{-x^4} \cdot (2x)$ $f'(x) = 2x \left( e^{-(x^4+2x^2+1)} - e^{-x^4} ight)$ $f'(x) = 2x \cdot e^{-x^4} \left( e^{-(2x^2+1)} - 1 ight)$ Since $e^{-x^4} > 0$ for all $x$,and $e^{-(2x^2+1)}  0$),the term $(e^{-(2x^2+1)} - 1)$ is always negative. For $f(x)$ to be an increasing function,we require $f'(x) \geq 0$. $2x \cdot (	ext{negative value}) \geq 0 \implies x \leq 0$. Thus,$f(x)$ is increasing on the interval $(-\infty, 0]$.

Let $f(x) = \int_{x^2}^{x^2+1} e^{-t^2} dt$,for $x \in (-\infty, \infty)$. For which interval is $f(x)$ an increasing function?

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