From Known Values to Guaranteed Range Values
A range question often gives only a few facts about a function: it is continuous on an interval, and it takes certain values at particular points. Those facts can be enough to establish that a target value occurs, even when the function is not monotone and no maximum or minimum is known to exist. The key is that the range of a continuous function on an interval is itself an interval.
All subsets of \(\mathbb{R}\) below have the subspace topology inherited from the standard topology on \(\mathbb{R}\). An interval \(J\subseteq\mathbb{R}\) means a set containing every real number between any two of its points. The range of \(f:J\to\mathbb{R}\) is the set \(f(J)=\{f(x):x\in J\}\). Continuity is with respect to the subspace topology on \(J\).
This conclusion concerns values between values that the function actually takes. It does not assert that the range has a largest or smallest element, and it does not require the domain interval to be closed or bounded.
The Range-Value Principle
Proof. The interval \(J\) is connected, so its continuous image \(f(J)\) is connected. A connected subset of \(\mathbb{R}\) is an interval. Both \(f(u)\) and \(f(v)\) belong to \(f(J)\); therefore, by the definition of an interval, every real number between them belongs to \(f(J)\). This includes the endpoints, which are already values \(f(u)\) and \(f(v)\). For \(f(u)<c<f(v)\), the result gives \(c\in f(J)\), meaning there is an \(x\in J\) with \(f(x)=c\). \(\square\)
The same conclusion holds if \(f(v)\leq f(u)\), by exchanging the roles of \(u\) and \(v\). If the two known values are equal, the interval between them consists of that single value, which is already attained. Thus the useful general rule is: every value between two attained values is attained.
For a zero question, the most useful case is often a sign change. If \(f(u)<0<f(v)\), then \(0\) lies strictly between the two known values, so \(f\) has a zero somewhere in \(J\). More generally, to prove that a target \(c\) is attained, it is enough to find two points whose function values lie on opposite sides of \(c\). There is no need to determine the whole range or locate an extremum.
Worked Example: A Target Value for a Polynomial
Let \(f:(0,\infty)\to\mathbb{R}\) be \(f(x)=x^2-3x\). Prove that \(f(x)=-1\) for some \(x>0\), using two known values, and check the conclusion directly.
A polynomial is continuous on \(\mathbb{R}\), so its restriction to the interval \((0,\infty)\) is continuous. At the points \(1\) and \(3\), both of which belong to the domain, the values are \(f(1)=1^2-3(1)=1-3=-2\) and \(f(3)=3^2-3(3)=9-9=0\). Since \(-2<-1<0\), the range-value principle guarantees a point \(x\in(0,\infty)\) with \(f(x)=-1\).
Direct check. Solving \(f(x)=-1\) gives \(x^2-3x=-1\), hence \(x^2-3x+1=0\). The quadratic formula gives \(x=(3-\sqrt{5})/2\) or \(x=(3+\sqrt{5})/2\). Both are positive because \(3>\sqrt{5}\), and each satisfies the equation by the quadratic formula. In particular, at least one solution lies in the domain. Substitution confirms that either solution has \(x^2-3x=-1\).
Finding an Exact Range Without Attained Endpoints
The range-value principle can do more than certify one target. If the range is an interval, bounds and values approaching the bounds can determine the entire range. The relevant distinction is between a bound and an attained value: a range can have an infimum or supremum that does not belong to it.
Proof. Because \(S\subseteq[m,M]\) and \(m,M\notin S\), every element of \(S\) lies in \((m,M)\), so \(S\subseteq(m,M)\). Now take any \(y\in(m,M)\). By hypothesis there are \(s,t\in S\) with \(s<y<t\). Since \(S\) is an interval and \(s,t\in S\), it contains every number between \(s\) and \(t\), including \(y\). Thus \((m,M)\subseteq S\). The two inclusions give \(S=(m,M)\). \(\square\)
In applications, one often has a simple way to produce the bracketing values. For each target \(y\) strictly between the proposed endpoints, find one input with output below \(y\) and another with output above \(y\). The two inputs may depend on \(y\); they do not have to be fixed for every target.
Worked Example: A Range with Neither Endpoint Attained
Find the range of \(f:(0,\infty)\to\mathbb{R}\) given by \(f(x)=\dfrac{x}{1+x}\).
The denominator \(1+x\) is positive for \(x>0\), and the quotient of continuous functions with nonzero denominator is continuous. For every \(x>0\), \(x>0\) gives \(f(x)>0\), while \(x<1+x\) gives \(f(x)<1\). Thus \(f((0,\infty))\subseteq(0,1)\), so neither endpoint is attained.
Fix any \(y\in(0,1)\). Choose a positive integer \(n\) large enough that \(n+1>1/y\) and \(n>y/(1-y)\). Such an integer exists by the Archimedean property. At the input \(1/n>0\), \(f(1/n)=(1/n)/(1+1/n)=1/(n+1)<y\), where the last inequality follows from \(n+1>1/y\). At the input \(n>0\), \(f(n)=n/(1+n)>y\), because this inequality is equivalent, after multiplying by \(1+n>0\), to \(n>y(1+n)\), or \(n(1-y)>y\), which follows from \(n>y/(1-y)\).
The range is an interval because \(f\) is continuous on the interval \((0,\infty)\). For each \(y\in(0,1)\), the two computed outputs lie on opposite sides of \(y\), so the range contains \(y\). Together with the strict bounds, this proves \(f((0,\infty))=(0,1)\).
Check with a specific target. For \(y=2/3\), the inputs \(1/2\) and \(3\) give \(f(1/2)=1/3<2/3\) and \(f(3)=3/4>2/3\). The value is indeed attained: solving \(x/(1+x)=2/3\) gives \(3x=2+2x\), hence \(x=2\), and \(f(2)=2/3\).
Worked Example: An Unbounded Range with a Missing Lower Endpoint
Find the range of \(g:\mathbb{R}\to\mathbb{R}\) defined by \(g(x)=e^x\), using connectedness and values on either side of a target.
The exponential function is continuous, and \(\mathbb{R}\) is an interval, so \(g(\mathbb{R})\) is an interval. Also \(e^x>0\) for every real \(x\), so the range is contained in \((0,\infty)\).
Take any target \(y>0\). The standard limits of the exponential function give \(e^{-n}\to0\) and \(e^n\to\infty\) as \(n\to\infty\). Hence there is a positive integer \(n\) large enough that \(e^{-n}<y<e^n\). Both values are in the range, as they are \(g(-n)\) and \(g(n)\). Since the range is an interval, it contains every value between them, in particular \(y\). This holds for every \(y>0\), so \(g(\mathbb{R})=(0,\infty)\).
Check with a particular target. For \(y=1\), the range already contains the target because \(g(0)=e^0=1\). For a target not given immediately by this input, such as \(y=2\), take any sufficiently large \(n\) with \(e^{-n}<2<e^n\). The interval property then guarantees that \(2\) is in the range; indeed \(g(\ln 2)=e^{\ln 2}=2\). The endpoint \(0\) is not attained because \(e^x>0\), even though values of the function approach \(0\).
Worked Example: A Nonmonotone Function Still Fills a Range Interval
Let \(h:[-2,2]\to\mathbb{R}\) be \(h(x)=x^3-3x\). Use selected values to show that every \(y\in[-2,2]\) is attained.
The function is a polynomial, hence continuous, and \([-2,2]\) is an interval. Direct substitution gives \(h(-2)=(-2)^3-3(-2)=-8+6=-2\) and \(h(2)=2^3-3(2)=8-6=2\). The range-value principle therefore gives \([-2,2]\subseteq h([-2,2])\). For a check at \(y=1\), the principle guarantees an input between \(-2\) and \(2\) with output \(1\). In fact \(h(-1)=(-1)^3-3(-1)=-1+3=2\) and \(h(0)=0\); these values bracket \(1\), and direct solution \(h(x)=1\) is not needed for the guarantee.
This example uses only continuity and two attained values. It does not require the function to be monotone: for instance, \(h(-1)=2\), \(h(0)=0\), and \(h(1)=1-3=-2\), so its values rise and fall across the domain. The range-value conclusion remains valid.
A Common Pitfall: Treating a Bound as an Attained Value
A familiar theorem says that a continuous real-valued function on a compact interval attains a maximum and a minimum. Its compactness hypothesis matters. On an open or unbounded interval, continuity alone does not guarantee either endpoint value is attained.
Wrong argument. Let \(f(x)=x\) on \((0,1)\). Since the values are less than \(1\), one might write “the maximum is \(1\), attained at \(x=1\).” This calculation uses an input that is not in the domain: \(1\notin(0,1)\). The values \(f(x)\) are all strictly below \(1\), and values can be arbitrarily close to \(1\), but there is no maximum.
Correction. For every \(x\in(0,1)\), \(0<f(x)<1\). Given any \(y\in(0,1)\), the input \(x=y\) belongs to the domain and satisfies \(f(x)=y\). Therefore the range is exactly \((0,1)\): its infimum is \(0\), its supremum is \(1\), and neither is attained. The connectedness method makes the distinction systematic: first establish which values are in the range, then separately check whether proposed endpoints are attained.
For range questions, a reliable order of work is:
- Check that the domain is an interval and the function is continuous there.
- Compute function values at inputs that are actually in the domain.
- Use values on opposite sides of a target to establish that the target is attained.
- If seeking the exact range, prove bounds and treat endpoint attainment separately.
Check Your Understanding
For each question, identify the interval and the known values or bounds that justify the range conclusion.
- If \(f:[-1,3]\to\mathbb{R}\) is continuous, \(f(-1)=4\), and \(f(3)=-2\), which interval of values must belong to \(f([-1,3])\)?
- Suppose \(f:(0,\infty)\to\mathbb{R}\) is continuous and \(f(1)=-3\), \(f(4)=5\). Must \(f(x)=0\) have a solution? Explain which result applies.
- For \(p(x)=x/(2+x)\) on \((0,\infty)\), prove from bounds and bracketing values that the range is \((0,1)\).
- Give an example of a continuous function on an interval whose supremum is not attained, and explain why the extreme value theorem does not apply.
- Why do two attained values \(f(u)\) and \(f(v)\) guarantee every value between them even when \(f\) is not monotone?