From continuity to an interval of outputs
A continuous function can rise, fall, or change direction. But on an interval of inputs, it cannot jump from one output value to another while skipping all the values in between. This is a direct consequence of the Intermediate Value Theorem (IVT).
The key point is that the theorem applies between any two input points in the domain. Their outputs need not occur at the endpoints of the whole domain, and the function need not be increasing or decreasing.
The image is also called the range of the function on \(I\). The interval property concerns values in this image: if two values belong to \(f(I)\), every value between them must belong to \(f(I)\) too.
The continuous-image theorem
This theorem says that a continuous function on an interval has an interval as its image. It does not say that the image must be a closed interval, or that it must include its largest or smallest possible values. Those depend on the domain and the function.
The IVT used here says that if \(f\) is continuous on a closed interval \([a,b]\), then for every value \(y\) between \(f(a)\) and \(f(b)\), there is a point \(c\in[a,b]\) such that \(f(c)=y\). The closed interval in this statement will come from choosing two points of the domain.
Why the theorem follows from the IVT
Take any two outputs \(u,v\in f(I)\) with \(u<v\). Because they are attained, there are points \(x_1,x_2\in I\) such that \(f(x_1)=u\) and \(f(x_2)=v\). The input points might be in either order. Relabel them, if necessary, so that the left-hand input is called \(a\) and the right-hand input is called \(b\). Since \(I\) is an interval and \(a,b\in I\), the whole closed segment \([a,b]\) lies in \(I\). Continuity of \(f\) on \(I\) therefore gives continuity on this segment, so the IVT applies there.
Let \(u=f(x_1)\) and \(v=f(x_2)\), where \(u<v\).
Order the input points as \(a<b\). Since both lie in \(I\), the entire segment \([a,b]\) lies in \(I\).
The endpoint outputs are \(u\) and \(v\), in one order or the other. The IVT guarantees that every value between them occurs at some point of \([a,b]\).
More explicitly, choose any \(y\) with \(u\leq y\leq v\). The number \(y\) lies between \(f(a)\) and \(f(b)\), regardless of which endpoint has the smaller output. By the IVT, some \(c\in[a,b]\) satisfies \(f(c)=y\). Because \([a,b]\subseteq I\), this shows \(y\in f(I)\). Since this argument works for every pair of outputs and every value between them, \(f(I)\) is an interval.
Worked examples
Worked Example: A quadratic on a closed interval
Find the image of \(f(x)=x^2\) on \(I=[-1,2]\), and relate it to the theorem.
First, \(x^2\geq 0\) for every real \(x\), so no negative output occurs. Also, for \(x\in[-1,2]\), we have \(|x|\leq 2\), hence \(x^2\leq 4\). Thus every output lies in \([0,4]\).
The lower endpoint is attained at \(x=0\), since \(0\in I\) and \(f(0)=0^2=0\). The upper endpoint is attained at \(x=2\), since \(f(2)=2^2=4\). For any \(y\in[0,4]\), the number \(\sqrt y\) satisfies \(0\leq\sqrt y\leq2\), so \(\sqrt y\in I\), and \(f(\sqrt y)=(\sqrt y)^2=y\).
Therefore every value in \([0,4]\) is attained, and no value outside it is attained. The image is exactly \([0,4]\). In particular, the IVT applied between the inputs \(0\) and \(2\) guarantees every output between \(f(0)=0\) and \(f(2)=4\).
Check: The proposed input \(\sqrt y\) is in the domain, and substituting it into \(f\) returns \(y\), as required.
Worked Example: A function that changes direction
Find the image of \(g(x)=x^2-2x\) on \(I=[0,3]\). This function does not increase throughout the interval, but that does not prevent the theorem from applying.
Complete the square: \(g(x)=x^2-2x=(x-1)^2-1\). Since a square is nonnegative, \(g(x)\geq-1\), with equality at \(x=1\). To establish an upper bound, compute \(g(x)-3=x^2-2x-3=(x-3)(x+1)\). For \(0\leq x\leq3\), \(x-3\leq0\) and \(x+1>0\), so \((x-3)(x+1)\leq0\), which gives \(g(x)\leq3\).
Both bounds are attained: \(g(1)=(1-1)^2-1=-1\), and \(g(3)=3^2-2(3)=9-6=3\). Now apply the IVT on \([1,3]\), where \(g\) is continuous and the endpoint outputs are \(-1\) and \(3\). Every \(y\in[-1,3]\) lies between those outputs, so the IVT guarantees an input \(c\in[1,3]\) with \(g(c)=y\). Together with the bounds, this proves that the image on \([0,3]\) is exactly \([-1,3]\).
Check: The lower-bound equality occurs at \(x=1\), and the upper-bound equality occurs at \(x=3\). Both points belong to \([0,3]\), and substituting them gives the claimed outputs.
Worked Example: An open domain and an excluded endpoint value
Find the image of \(h(x)=x^2\) on the open interval \(I=(-2,1)\).
Every output is nonnegative. Also, if \(-2<x<1\), then \(|x|<2\), so \(x^2<4\). Therefore \(h(I)\subseteq[0,4)\).
Conversely, take any \(y\in[0,4)\). Then \(0\leq\sqrt y<2\), so \(-2<-\sqrt y\leq0\). In particular, \(-\sqrt y\in(-2,1)\). Substitution gives \(h(-\sqrt y)=(-\sqrt y)^2=y\). Thus every \(y\in[0,4)\) is attained, and the image is exactly \([0,4)\).
The output \(0\) is included because \(h(0)=0\), but \(4\) is not included: achieving \(x^2=4\) requires \(x=-2\) or \(x=2\), neither of which belongs to \((-2,1)\). The image is still an interval even though it is open at one end and closed at the other.
Check: For every proposed \(y\in[0,4)\), the chosen input \(-\sqrt y\) lies in the domain and its square is \(y\). The excluded value \(4\) has no permitted input.
What the interval conclusion does—and does not—say
The theorem is a guarantee about values between outputs, not a formula for finding the full image. In the examples, additional bounds or explicit choices of inputs helped identify the exact image. The IVT itself supplies the no-gaps conclusion: once two outputs are known to occur, all the values between them occur as well.
Nor does the theorem require the domain to be closed or bounded. Its proof only needs that the domain is an interval, that the two chosen input points belong to it, and that the function is continuous there. For those two inputs, the interval property puts the closed segment between them inside the domain. The IVT can then be applied on that segment.
A common mistake: ignoring the domain
Continuity alone does not guarantee an interval of outputs if the domain itself is not an interval. For example, let \(D=[-1,0]\cup[1,2]\), and define \(q:D\to\mathbb{R}\) by \(q(x)=x\). This function is continuous on its domain, and its image is \([-1,0]\cup[1,2]\). The output \(0\) occurs at \(x=0\), and the output \(1\) occurs at \(x=1\), but the value \(1/2\) is not attained.
A mistaken argument might say, “The function is continuous between \(x=0\) and \(x=1\), so the IVT gives an input where \(q(x)=1/2\).” The problem is that \(q\) is not defined at the points between \(0\) and \(1\): the segment \([0,1]\) is not contained in \(D\). Thus the IVT does not apply on that segment. The correction is to check the domain before using the theorem. The continuous-image result requires an interval domain precisely so that the segment between any two chosen inputs is available.
Check Your Understanding
Use the continuous-image theorem and the IVT where appropriate.
- If \(f\) is continuous on an interval \(I\), and \(f(a)=2\), \(f(b)=7\) for \(a,b\in I\), what can you conclude about the values from \(2\) to \(7\)?
- Why can the theorem be applied to the segment between two points of an interval, even if the whole domain is open or unbounded?
- For \(f(x)=x^2\) on \((-2,1)\), why is \(4\) not in the image, while \(0\) is?
- Does a continuous function on a domain that is not an interval necessarily have an interval as its image? Explain using \(q(x)=x\) on \([-1,0]\cup[1,2]\).
- Does the continuous-image theorem say that the image must include its supremum and infimum? Explain using one of the examples.
