Poemslore8ye

2022-05-02

Antiderivative of a branch function

Let $f:\mathbb{R}\mapsto \mathbb{R}$, $f(x)={e}^{x}$ for $x\le 0$ and $ax+b$ for $x>0$. I want to determine the real numbers a and b for which f admits antiderivatives.

I took a plausible antiderivative $F(x)={e}^{x}+c$ for $x<0$, $F(0)=\alpha $ and $a\frac{{x}^{2}}{2}+bx+d$ for $x>0$ and using the fact that F must be first continuous and then differentiable, I showed that $b=1$.

This is easy to see as from continuity we must have that $d=\alpha =c+1$.

Then, since F is clearly differentiable on $\mathbb{R}\setminus \{0\}$ if we impose the condition that the left and right derivatives at 0 of F to be equal so we must have:

$\underset{x\to 0,x<0}{lim}\frac{F(x)-F(0)}{x-0}=\underset{x\to 0,x>0}{lim}\frac{F(x)-F(0)}{x-0}$

Thus: $\underset{x\to 0,x<0}{lim}\frac{{e}^{x}+c-c-1}{x}=\underset{x\to 0,x>0}{lim}\frac{a\frac{{x}^{2}}{2}+bx+c+1-c-1}{x}$

Which gives us $b=1$.

My question is then, does this suffice? Can a be chosen arbitrarily such that f has antiderivatives as long as $b=1$. It would seem that the function $F(X)={e}^{x}+c$ for $x\le 0$ and $F(x)=a\frac{{x}^{2}}{2}+x+c+1$ for $x>0$ for $c\in \mathbb{R}$ does indeed satisfy the necessary conditions.

Let $f:\mathbb{R}\mapsto \mathbb{R}$, $f(x)={e}^{x}$ for $x\le 0$ and $ax+b$ for $x>0$. I want to determine the real numbers a and b for which f admits antiderivatives.

I took a plausible antiderivative $F(x)={e}^{x}+c$ for $x<0$, $F(0)=\alpha $ and $a\frac{{x}^{2}}{2}+bx+d$ for $x>0$ and using the fact that F must be first continuous and then differentiable, I showed that $b=1$.

This is easy to see as from continuity we must have that $d=\alpha =c+1$.

Then, since F is clearly differentiable on $\mathbb{R}\setminus \{0\}$ if we impose the condition that the left and right derivatives at 0 of F to be equal so we must have:

$\underset{x\to 0,x<0}{lim}\frac{F(x)-F(0)}{x-0}=\underset{x\to 0,x>0}{lim}\frac{F(x)-F(0)}{x-0}$

Thus: $\underset{x\to 0,x<0}{lim}\frac{{e}^{x}+c-c-1}{x}=\underset{x\to 0,x>0}{lim}\frac{a\frac{{x}^{2}}{2}+bx+c+1-c-1}{x}$

Which gives us $b=1$.

My question is then, does this suffice? Can a be chosen arbitrarily such that f has antiderivatives as long as $b=1$. It would seem that the function $F(X)={e}^{x}+c$ for $x\le 0$ and $F(x)=a\frac{{x}^{2}}{2}+x+c+1$ for $x>0$ for $c\in \mathbb{R}$ does indeed satisfy the necessary conditions.

veceritzpzg

Beginner2022-05-03Added 16 answers

Explanation:

By definition, f admits an antiderivative iff f is a derivative of some function. Thus, by Darboux theorem, f has to have an intermediate value property. For f of the given form this is equivalent to $b=1$ as you got as well. Then, independently of the choice of a, f is continuous and has an antiderivative by (one of) the Fundamental Theorem(s) of Calculus.

By definition, f admits an antiderivative iff f is a derivative of some function. Thus, by Darboux theorem, f has to have an intermediate value property. For f of the given form this is equivalent to $b=1$ as you got as well. Then, independently of the choice of a, f is continuous and has an antiderivative by (one of) the Fundamental Theorem(s) of Calculus.

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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