Wemazi

Topic 5.1

Check when the theorem applies

Builds on checking the conditions of a rule such as L'Hôpital's rule. 7 minutes.

A secant line and a tangent

The curve is the graph of \(f(x) = \tfrac12 x^2 + 1\) on the interval \([0, 4]\). The solid straight line joins the two end points of the curve, \((0, 1)\) and \((4, 9)\), and it is called the secant line.

The secant line rises 8 units over 4 units, so its slope is 2. This number is the average rate of change of \(f\) on \([0, 4]\).

Drag the orange point along the curve. The orange line is the tangent at the point, and the dashed line through the point has the same slope as the secant line.

At \(x = 2\), the tangent lies along the dashed line, so it is parallel to the secant line and its slope is also 2. At every other point, the tangent is flatter or steeper.

Any smooth curve

The numbers are gone. The curve has no break and a tangent at every point, and the secant line joins the points of the curve above \(a\) and \(b\).

Move the slider for b. The secant line turns, because its right end moves along the curve.

Move the orange point until its tangent lies along the dashed line. For every \(b\), the curve has such a point between \(a\) and \(b\).

Notation

  1. The secant line joins \((a, f(a))\) and \((b, f(b))\). Its slope is the average rate of change of \(f\) on \([a, b]\).

    \[\frac{f(b) - f(a)}{b - a}\]

  2. The first condition is that \(f\) is continuous on the closed interval \([a, b]\). This means that the graph has no gap and no jump from \(a\) to \(b\), and that \(f\) has a value at both end points.

  3. The second condition is that \(f\) is differentiable on the open interval \((a, b)\). This means that every point strictly between \(a\) and \(b\) has a tangent, so the graph has no corner or vertical tangent there. The end points need no derivative.

  4. If both conditions hold, the tangent at some number \(\cy{c}\) between \(a\) and \(b\) is parallel to the secant line. The two slopes are then equal.

    \[\cy{f'(c)} = \frac{f(b) - f(a)}{b - a}\]

    This is the mean value theorem. It guarantees that \(c\) exists, but not where \(c\) is.

When it fails

Each condition can fail. With the jump, \(f\) is not continuous on \([a, b]\). The secant line falls while every tangent rises, so none is parallel to it.

With the corner, \(f'\) does not exist there. Tangents left of the corner have a larger slope than the secant line, and those right of it a smaller slope. None is parallel to it.

Example 1: a square root

Does the theorem apply to \(f(x) = \sqrt{x}\) on \([0, 9]\)?

  1. The square root has a value at every \(x \ge 0\) and no gap, so \(f\) is continuous on \([0, 9]\).

  2. For \(x > 0\), the derivative is a number, so every point of \((0, 9)\) has a tangent.

    \[f'(x) = \dfrac{1}{2\sqrt{x}}\]

  3. At \(x = 0\), the tangent is vertical, so \(f'(0)\) does not exist. The point 0 is an end point, so the second condition still holds.

  4. Both conditions hold, so the theorem applies. Some tangent has the slope of the secant line, \(\tfrac13\).

    \[\cy{\dfrac{3 - 0}{9 - 0}} = \cy{\dfrac13}\]

    As a check, the tangent at \(x = \tfrac94\) has slope \(\dfrac{1}{2 \cdot \frac32} = \tfrac13\).

Example 2: a graph with a break

Decide whether the theorem applies to \(f(x) = \dfrac{1}{x - 2}\) on \([0, 5]\).

  1. The secant line joins \((0, -\tfrac12)\) and \((5, \tfrac13)\), so its slope is \(\tfrac16\).

    \[\dfrac{f(5) - f(0)}{5 - 0} = \dfrac{\frac13 + \frac12}{5} = \dfrac16\]

  2. A tempting wrong move: The function is a quotient of polynomials and has a derivative formula, so both conditions hold.

    \[f'(x) = -\dfrac{1}{(x - 2)^2}\]

    The graph has a vertical asymptote at \(x = 2\), inside \([0, 5]\), and it is cut in two. Every tangent slopes down while the secant line slopes up, so none is parallel to it.

  3. The function has no value at \(x = 2\), and 2 lies in \([0, 5]\). This means that \(f\) is not continuous on \([0, 5]\), so the theorem does not apply.

Practice

Decide whether the theorem applies to \(f(x) = \ln x\) on \([0, 3]\).

Follow the curve toward \(x = 0\). Does \(f\) have a value there?

Solution

  1. The function \(\ln x\) is defined only for \(x > 0\), so \(f(0)\) does not exist.

  2. The first condition needs a value at the end point 0, so the theorem does not apply on \([0, 3]\).

  3. On \([1, 3]\), the theorem applies, because \(f\) is continuous there and \(f'\) exists on \((1, 3)\).

    \[f'(x) = \dfrac{1}{x}\]

Questions

The graph shows \(g\) from \(x = 0\) to \(x = 9\), with a corner at \(x = 3\) and a jump at \(x = 6\). At the jump, the filled dot lies above the open ring.

  1. On which interval do both conditions of the theorem hold for \(g\)?

    • \([0, 3]\)
    • \([1, 5]\)
    • \([4, 8]\)
    Answer

    \([0, 3]\)

  2. Why does the theorem not apply to \(g\) on \([2, 5]\)?

    • \(g\) has a corner at \(x = 3\), so it is not differentiable on \((2, 5)\).
    • \(g\) is not continuous at \(x = 3\).
    • \(g\) is not differentiable at the end point \(x = 2\).
    Answer

    \(g\) has a corner at \(x = 3\), so it is not differentiable on \((2, 5)\).

  3. Which function satisfies both conditions of the theorem on \([-1, 4]\)?

    • \(\dfrac{1}{x - 6}\)
    • \(|x - 2|\)
    • \(\dfrac{1}{x - 1}\)
    Answer

    \(\dfrac{1}{x - 6}\)

Summary

You check that the function is continuous on the closed interval and differentiable on the open interval.

This matters because a parallel tangent needs a connected curve with a tangent at every point between the end points.

You do this before you use the theorem, whenever a problem asks for a guarantee.

This is one of five skills in topic 5.1 of Which & Why: AP Calculus.