Skip to main content

Subsection 2.13.2 Back to the MVT

Rolle's theorem can be generalised in a straight-forward way; given a differentiable function \(f(x)\) we can still say something about \(\diff{f}{x}\text{,}\) even if \(f(a) \neq f(b)\text{.}\) Consider the following sketch:

Figure 2.13.4

All we have done is tilt the picture so that \(f(a) \lt f (b)\text{.}\) Now we can no longer guarantee that there will be a point on the graph where the tangent line is horizontal, but there will be a point where the tangent line is parallel to the secant joining \((a, f(a))\) to \((b, f(b))\text{.}\)

To state this in terms of our first scenario back at the beginning of this section, suppose that you are driving along the \(x\)–axis. At time \(t=a\) you are at \(x=f(a)\) and at time \(t=b\) you are at \(x=f(b)\text{.}\) For simplicity, let's suppose that \(b \gt a\) and \(f(b)\ge f(a)\text{,}\) just like in the above sketch. Then during the time interval in question you travelled a net distance of \(f(b)-f(a)\text{.}\) It took you \(b-a\) units of time to travel that distance, so your average velocity was \(\frac{f(b)-f(a)}{b-a}\text{.}\) You may very well have been going faster than \(\frac{f(b)-f(a)}{b-a}\) part of the time and slower than \(\frac{f(b)-f(a)}{b-a}\) part of the time. But it is reasonable to guess that at some time between \(t=a\) and \(t=b\) your instantaneous velocity was exactly \(\frac{f(b)-f(a)}{b-a}\text{.}\) The mean value theorem says that, under reasonable assumptions about \(f\text{,}\) this is indeed the case.

Let us start to explore the mean value theorem — which is very frequently known as the MVT. A simple example to start:

Consider the polynomial \(f(x)=3x^2-4x+2\) on \([-1,1]\text{.}\)

  • Since \(f\) is a polynomial it is continuous on the interval and also differentiable on the interval. Hence we can apply the MVT.
  • The MVT tells us that there is a point \(c \in (-1,1)\) so that
    \begin{align*} f'(c) &= \frac{f(1)-f(-1)}{1-(-1)} = \frac{1-9}{2} =-4 \end{align*}

This example is sufficiently simple that we can find the point \(c\) and the corresponding tangent line:

  • The derivative is
    \begin{align*} f'(x) &= 6x-4 \end{align*}
  • So we need to solve \(f'(c)=-4\text{:}\)
    \begin{align*} 6c-4 &= -4 \end{align*}
    which tells us that \(c=0\text{.}\)
  • The tangent line has slope \(-4\) and passes through \((0,f(0))=(0,2)\text{,}\) and so is given by
    \begin{align*} y &= -4x+2 \end{align*}
  • The secant line joining \((-1,f(-1))=(-1,9)\) to \((1,f(1))=(1,1)\) is just
    \begin{align*} y &= 5-4x \end{align*}

  • Here is a sketch of the curve and the two lines:

We can return to our initial car-motivated examples. Say you are driving along a straight road in a car that can go at most \(80km/h\text{.}\) How far can you go in 2 hours? — the answer is easy, but we can also solve this using MVT.

  • Let \(s(t)\) be the position of the car in \(km\) at time \(t\) measured in hours.
  • Then \(s(0)=0\) and \(s(2)=q\text{,}\) where \(q\) is the quantity that we need to bound.
  • We are told that \(| s'(t) | \leq 80\text{,}\) or equivalently
    \begin{gather*} -80 \leq s'(t) \leq 80 \end{gather*}
  • By the MVT there is some \(c\) between 0 and 2 so that
    \begin{align*} s'(c) &= \frac{q-0}{2} = \frac{q}{2} \end{align*}
  • Now since \(-80 \leq s'(c) \leq 80\) we must have \(-80 \leq q/2 \leq 80\) and hence \(-160 \leq q=s(2) \leq 160\text{.}\)

More generally if we have some information about the derivative, then we can use the MVT to leverage this information to tell us something about the function.

Let \(f(x)\) be a differentiable function so that

\begin{align*} f(1)&=10 &\text{ and }&& -1 \leq f'(x) \leq 2 \text{ everywhere} \end{align*}

Obtain upper and lower bounds on \(f(5)\text{.}\)

Okay — what do we do?

  • Since \(f(x)\) is differentiable we can use the MVT.
  • Say \(f(5)=q\text{,}\) then the MVT tells us that there is some \(c\) between \(1\) and \(5\) such that
    \begin{align*} f'(c) &=\frac{q-10}{5-1} = \frac{q-10}{4} \end{align*}
  • But we know that \(-1 \leq f'(c) \leq 2\text{,}\) so
    \begin{align*} -1 &\leq f'(c) \leq 2\\ -1 & \leq \frac{q-10}{4} \leq 2\\ -4 & \leq q-10 \leq 8\\ 6 \leq q \leq 18 \end{align*}
  • Thus we must have \(6 \leq f(5) \leq 18\text{.}\)
login