Consider $x_0 \le x \le x_1$ and $y_0 \le y \le y_1$ . I'm looking for a formula that indicates interval of $xy$ . For example if we have $-1\le x\le 2 $ , how we can understand interval of $x^2 - 2x$ without graphing it . I think if there is a general formula , it can be very helpful .

  • $\begingroup$ You are saying $x^2-2x\leq 0$, isn't? $\endgroup$ – MAN-MADE Jul 8 '17 at 5:00
  • $\begingroup$ It seems that you want the minimum and the maximum of the function $f(x)=x^2−2x$ on the interval [-1,2]. Is this right? $\endgroup$ – miracle173 Jul 8 '17 at 6:47
  • $\begingroup$ @miracle173 No , I'm looking for a way for multiplying intervals . For finding extremum , we can use derivative easily . $\endgroup$ – S.H.W Jul 8 '17 at 10:45
  • $\begingroup$ @MANMAID you define $S_A$, but then don't use it in your question. That does not make sense. And why do you want $xf(x)$? What is $f$? $\endgroup$ – miracle173 Jul 8 '17 at 11:43
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    $\begingroup$ Your question is rather unclear to me but you tagged it interval-arithmetic. Did you alread read the wiki article of this topic? Is this the topic you are asking for? $\endgroup$ – miracle173 Jul 8 '17 at 11:48

Wikipedia says:

Evaluating a function at each element of a subset X of the domain, produces a set called the image of X under or through the function.

This image of a set $X$ under a function $f$ is usually written as $f(X)$ and is formally defined as

$$f(X)=\{f(x)|x\in X\}$$

So you want know the image of the interval $[-1,2]$ under the continuous function $x\to x^2-2x$ and the image of the rectangle $[x_0,x_1]\times [y_0,y_1]$ under the continuous function $(x,y)\to xy$.

Your phrase "interval of $x^2-2x$" is not used for this and may not be understood.

But it can be shown that


From this follows

  • If $f:[a,b]\mapsto \mathbb{R}$ is a continuous function, then the image of $[a,b]$ under $f$, $f([a,b])$, is an interval.
  • If $f:[a,b]\times[c,d]\mapsto \mathbb{R}$ is a continuous function, then the image of $[a,b]\times[c,d]$ under $f$, $f([a,b]\times[c,d])$, is an interval.

If the image $f(X)$ of $X$ under $f$ is an interval then we have $$f(X)=[\min(f(X)),\max(f(X))]$$

So to find the left and the right endpoint if the interval we have to find the minimum and the maximum of $f$ in $X$.

If f is a differentiable function from $[a,b]$ we know that its extremal values (= minimum or maximum values) are either on the endpoints $a$ or $b$ of the interval $X=[a,b]$ or at one of its local extremal points in the inner of $X$. These points can be found by using the derivative.

So for $$f(x)=x^2-2x$$ we have $$f'(x)=2x-2$$ which is $0$ for $x=1$.

So $$\min(f[-1,2])=\min\{f(-1),f(0),f(2)\}=\min\{3,-1,0\}=-1$$ and $$\max(f[-1,2])=\max\{f(-1),f(0),f(2)\}=\max\{3,-1,0\}=3$$

and therefore


For the two dimensional differential function $$p(x,y)=xy$$ similar statements hold. But even without differential calculus one can show that


Other functions can be found in this wiki article about interval arithmetic where intervals are used to find bounds for rounding errors in computations.

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