IVT, MVT and extreme values: check the assumptions first

Three questions, three different guarantees

Throughout, $a<b$ are finite real numbers. A function is continuous on $[a,b]$ when it is continuous inside the interval and has the appropriate one-sided continuity at each endpoint. Differentiability is required only where stated. A prime denotes a derivative.

• Intermediate Value Theorem (IVT): if $f$ is continuous on $[a,b]$, every output between $f(a)$ and $f(b)$ occurs somewhere in that interval. In particular, opposite endpoint signs give a zero strictly between the endpoints. This does not establish uniqueness.
• Mean Value Theorem (MVT): if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there is at least one $c\in(a,b)$ such that $$f'(c)=\frac{f(b)-f(a)}{b-a}.$$ A tangent slope matches the interval’s secant slope. The theorem does not say that $c$ is the midpoint or is unique.
• Extreme Value Theorem (EVT): a continuous function on the closed, bounded interval $[a,b]$ attains both an absolute maximum and an absolute minimum. It does not locate them and does not require differentiability.

Failure of an assumption means that theorem cannot provide its guarantee. It does not, by itself, prove that the conclusion is false. Look at the actual function before making that stronger statement.

Worked example 1: prove one root exists, then rule out a second

Show that $p(x)=x^3+x-1$ has exactly one real root, and place it between two numbers without using a cubic formula.

1. Establish existence on a named interval

A polynomial is continuous on $[0,1]$. Its endpoint values are $p(0)=-1$ and $p(1)=1$. Since zero lies between them, the IVT gives at least one root in $(0,1)$. This is an existence statement, not yet an exact count.

2. Use a derivative to establish uniqueness

For every real $x$, $p'(x)=3x^2+1>0$. To justify why this prevents two roots, take any $u<v$. The polynomial meets the MVT assumptions on $[u,v]$, so for some $c\in(u,v)$,

$$p(v)-p(u)=p'(c)(v-u)>0.$$

Thus $p$ is strictly increasing on the whole real line. Two different inputs cannot both produce zero. Combine “at least one” with “at most one” to conclude that exactly one real root exists.

3. Narrow the location without claiming an exact answer

$$p(0.68)=-0.005568,\qquad p(0.69)=0.018509.$$

Continuity gives a root between these inputs, and uniqueness identifies it as the same root. Therefore $0.68<r<0.69$. The midpoint $0.685$ is an approximation with absolute error less than $0.005$, not an exact solution. If you need a tighter estimate, bisect the interval again and keep the subinterval with opposite endpoint signs.

Mistake to avoid: saying “the IVT proves exactly one root.” The IVT supplied existence; the positive derivative supplied uniqueness.

Worked example 2: find the tangent that matches an average rate

For $f(x)=\ln x$ on $[1,4]$, verify that the MVT applies and find every point where its conclusion holds. Here $\ln$ is the natural logarithm.

1. Check the entire interval

The interval contains only positive inputs. Therefore $\ln x$ is continuous on $[1,4]$ and differentiable on $(1,4)$. Checking only the endpoint values would not establish these assumptions.

2. Calculate the average slope, then match it

$$m=\frac{\ln4-\ln1}{4-1}=\frac{\ln4}{3}.$$

Since $f'(x)=1/x$, a point satisfying the conclusion solves

$$\frac1c=\frac{\ln4}{3},\qquad c=\frac3{\ln4}\approx2.164.$$

This lies strictly between 1 and 4, as required. It is not the midpoint $2.5$. The derivative $1/x$ is strictly decreasing for positive $x$, so this particular example has exactly one such point. Uniqueness came from the derivative’s behaviour, not from the MVT itself.

3. Interpret the two lines

The secant line through the endpoints and the tangent at $c$ have the same slope $m$. The tangent is

$$y=\ln c+\frac1c(x-c).$$

They are parallel, but they are not the same line. The theorem matches rates, not the two lines’ vertical positions. A sketch should show the tangent touching at an interior input, rather than automatically at the midpoint.

Four graduated checks: choose the claim before the calculation

Keep each solution closed until you have written the theorem’s assumptions and the precise conclusion you want.

1. A sign change across a missing input

Let $f(x)=1/(x-2)$, defined for $x\ne2$. Since $f(1)=-1$ and $f(3)=1$, a student claims there is a zero in $(1,3)$. Diagnose the claim. Does a zero actually exist?

2. Equal endpoint heights, but a corner inside

Let $g(x)=|x-2|$ on $[0,4]$. Its average slope is zero. Can you use the MVT to conclude that $g'(c)=0$ for some interior $c$? Check the actual derivative.

3. Attained extremes versus limiting values

Consider $h(x)=(x-2)^2$. Find its absolute extrema first on $[1,4]$, then on $(1,4)$. Which conclusion is guaranteed by the EVT?

4. Combine existence, uniqueness and a derivative bound

Let $q(x)=e^x+x-3$. Prove that it has exactly one real zero $r$, and that $0<r<1$. Then use the MVT to show $$0<1-r<\frac{e-2}{2}.$$ Explain why this inequality estimates location without solving for $r$.

What to practise next

For each argument, underline the interval, circle the assumptions and label the result as existence, uniqueness, location or attainment. If the issue is a cancelled factor or a missing function value, revisit domains and limits. If you need help applying the reasoning to your assigned material, return to MATH 140 tutoring.

Standard theorem references: OpenStax Calculus Volume 1 on continuity, the Mean Value Theorem and maxima and minima. Exercises and explanations above are original.