find the range of $x$ on which $f$ is decreasing, where $f(x)=int_0^{x^2-x}e^{t^2-1}dt$

Mathematics Asked by Steven Lu on December 4, 2020

I want to find the range of $x$ on which $f$ is decreasing, where

Let $u=x^2-x$, then $frac{du}{dx}=2x-1$, then $$f'(x)=frac{d}{dx}int_0^{x^2-x}e^{t^2-1}dt=frac{du}{dx}frac{d}{du}int_0^{x^2-x}e^{t^2-1}dt=(2x-1)e^{x^4-2x^3+x^2-1}$$

Since $e^{x^4-2x^3+x^2-1}>0$ for all $xin Bbb R$ and $2x-1<0iff x<frac{1}{2}$. $f$ is decreasing on $(-infty,frac{1}{2})$.

Furthermore, $f$ is increasing on $(frac{1}{2},infty)$, $f$ is differentiable at $x=frac{1}{2}$, and $f'(frac{1}{2})=0$, $f$ attains its minimum value at $x=frac{1}{2}$.

Am I right?

2 Answers

$f(x)=int_{0}^{x^2-x} e^{t^2-1} dt implies f'(x)= (2x-1) e^{(x^2-x)^2-1} >0 ~if~ x>1/2$. Hence $f(x)$ in increasing for $x>1/2$ and decreasing for $x<1/2$. Yes you are right. there is a min at $x=1/2$. This one point does not matter, you may also say that $f(x)$ is increasing in $[1/2,infty)]$ and decreasing on 4(-infty, 1/2]$.

Note: whether a function increasing or decreasing is decided by two points (not one). For instance, $x_1>x_2 leftrightarrows f(x_1) > f(x_2).$ If $f(x)$ is decreasing.

Answered by Z Ahmed on December 4, 2020

Everything is fine ! A little bit more can be said:

$f$ is strictly decreasing on $(-infty,frac{1}{2}]$


$f$ is strictly increasing on $[frac{1}{2}, infty).$

Answered by Fred on December 4, 2020

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