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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 .

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  • $\begingroup$ 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
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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

and

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

$$f([-1,2])=[-1,3]$$

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

$$p([x_0,x_1]\times[y_0,y_1])=[\min\{x_0y_0,x_0y_1,x_1y_0,x_1y_1\},\max\{x_0y_0,x_0y_1,x_1y_0,x_1y_1\}]$$

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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