Wemazi

Topic 1.11

Find a parameter that makes a join continuous

Builds on checking that the pieces of a piecewise function meet at a join. 7 minutes.

A constant that moves one piece

The left piece is the line \(y = kx + 1\) for \(x < 2\). Its slope \(k\) is a constant, which means a fixed number that you may choose.

The right piece is \(y = 9 - x^2\) for \(x \ge 2\). Its height at \(x = 2\) is \(9 - 4 = 5\), and the filled dot marks this value.

Drag the slider for \(k\). The left piece turns about the point \((0, 1)\), so its end at \(x = 2\), the open orange ring, moves up and down. The right piece stays in place.

The left end has the height \(2k + 1\), and the red bar shows the gap between this height and 5. When \(k = 2\), the gap is 0, so the pieces meet and the function is continuous at 2.

One equation closes the gap

The numbers are gone. The left piece depends on \(k\), and the right piece does not.

Drag the slider. The left end moves along the faint vertical line through the join, and the red bar changes length.

The gap closes at one position of the left end, where the left height equals the right height. This condition is an equation in k, and solving it gives the constant.

Notation

  1. A parameter is a constant that is written as a letter, such as \(k\). The problem asks for the value of \(k\) that makes the function continuous at the join \(c\).

  2. Write the height of the left piece at \(c\). It contains \(k\), so the left end moves when \(k\) changes.

    \[\cy{L(k)}\]

  3. Write the height of the right piece at \(c\). It may also contain \(k\). Here it does not, so the dot stays where it is.

    \[\cy{R}\]

  4. Continuity requires the two heights to be equal. This gives one equation in \(k\).

    \[\cy{L(k)} = \cy{R}\]

    The value \(f(c)\) is the height of the piece whose rule includes \(c\). The equation makes the other piece match it.

  5. Solve for k. Then substitute the solution into both pieces to check that the two heights are equal.

When it fails

Suppose the join is at \(x = 0\), the left piece is \(y = kx\), and the right piece is \(y = x + 3\). Press Pin the left end, then drag the slider for \(k\).

At \(x = 0\), the left height is \(k \cdot 0 = 0\) for every \(k\), so the piece turns about its end and the end stays below the dot. The equation \(0 = 3\) has no solution, so no value of \(k\) makes the function continuous.

Example 1: one constant in the left piece

Let \(f(x) = kx - 1\) for \(x < 3\), and let \(f(x) = 2x + 5\) for \(x \ge 3\). Find \(k\) so that \(f\) is continuous at \(x = 3\).

  1. At \(x = 3\), the left piece reaches the height \(3k - 1\).

    \[L(k) = \cy{3k - 1}\]

  2. The right piece has the height \(2(3) + 5 = 11\) at \(x = 3\). This is also the value \(f(3)\), because the rule \(x \ge 3\) includes 3.

    \[R = \cy{11}\]

  3. Set the two heights equal and solve for k.

    \[3k - 1 = 11 \Rightarrow 3k = 12 \Rightarrow k = 4\]

  4. Substitute \(k = 4\). The left height is \(4(3) - 1 = 11\), and the right height is 11, so the red bar is gone.

Example 2: a constant in both pieces

Let \(g(x) = kx^2 + 1\) for \(x < 2\), and let \(g(x) = 7 - kx\) for \(x \ge 2\). Find \(k\) so that \(g\) is continuous at \(x = 2\).

  1. Replace \(x\) by 2 in the left formula. The left piece reaches the height \(4k + 1\).

    \[L(k) = \cy{4k + 1}\]

  2. A tempting wrong move: The right piece is 7 at the join, so \(4k + 1 = 7\), and \(k = \tfrac{3}{2}\).

    \[4k + 1 = 7\]

    The right formula also contains k. With k = 3/2, the right height is 7 − 3 = 4, not 7. A red bar of length 3 remains between the left end at 7 and the right piece at 4.

  3. Replace \(x\) by 2 in the right formula as well. The right piece has the height \(7 - 2k\). Set the two heights equal.

    \[\begin{aligned} 4k + 1 &= 7 - 2k \\ 6k &= 6 \\ k &= 1 \end{aligned}\]

  4. Substitute \(k = 1\). The left height is \(4 + 1 = 5\), and the right height is \(7 - 2 = 5\), so the red bar is gone.

Practice

Let \(h(x) = 4x - k\) for \(x < 1\), and let \(h(x) = kx^2 + 2\) for \(x \ge 1\). Find \(k\) so that \(h\) is continuous at \(x = 1\).

Both pieces contain \(k\). Write the height of each piece at \(x = 1\). Then close the red bar in the picture.

Solution

  1. The left height at \(x = 1\) is \(4 - k\), and the right height is \(k + 2\).

    \[L(k) = \cy{4 - k}, \quad R(k) = \cy{k + 2}\]

  2. Set the heights equal and solve.

    \[4 - k = k + 2 \Rightarrow 2k = 2 \Rightarrow k = 1\]

  3. Substitute \(k = 1\). The left height is 3, and the right height is \(1 + 2 = 3\). The pieces meet at the dot.

Questions

Use the graph for the first two questions. The left piece is \(y = kx^2 - 4\) for \(x < 3\), drawn with \(k = \tfrac{1}{3}\). The right piece is \(y = x + 2\) for \(x \ge 3\).

  1. What height does the left piece reach at \(x = 3\) when \(k = \tfrac{1}{3}\)?

    • \(-1\)
    • 5
    • \(-4\)
    • 6
    Answer

    \(-1\)

  2. The slider increases \(k\). What happens in the picture?

    • The left end rises toward the dot, and the right piece stays in place.
    • Both pieces rise.
    • The dot moves down toward the left end.
    Answer

    The left end rises toward the dot, and the right piece stays in place.

  3. Let \(m(x) = 2kx - 1\) for \(x < 2\), and let \(m(x) = x^2 + k\) for \(x \ge 2\). Which \(k\) makes \(m\) continuous at \(x = 2\)?

    • \(\tfrac{5}{3}\)
    • \(\tfrac{5}{4}\)
    • 1
    • \(\tfrac{3}{5}\)
    Answer

    \(\tfrac{5}{3}\)

Summary

You write the height of each piece at the join, set the two heights equal, and solve for the constant.

This works because a piecewise function is continuous at a join when the pieces meet, and the equation says that they meet.

Use it when a piecewise formula contains an unknown constant and the problem asks for continuity.

This is one of four skills in topic 1.11 of Which & Why: AP Calculus.