Wemazi

Topic 5.4

Report the requested location or point

Builds on classifying a candidate using the sign change. 6 minutes.

Three answers from one top

The graph shows a function f with a local maximum at the top. One top can answer three different questions.

The question "where does f have its local maximum?" asks for an input. Press the first button. The orange line drops to the x-axis at x = 2, so the answer is x = 2.

The question "what is the local maximum value?" asks for an output. Press the second button. The dashed orange line runs to the y-axis at height 3, so the answer is 3.

The question "what is the point?" asks for both numbers. Press the third button. The answer is (2, 3), the input first and the output second.

Input, output, point

The scene is the same without numbers. The orange dot is at the top. Its input \(c\) is marked on the horizontal axis, and its height \(f(c)\) on the vertical axis.

The location of the local maximum is \(c\), the value is \(f(c)\), and the point is \((c, f(c))\), both numbers together on the curve.

Press the buttons to move the dot between the top and the bottom. The labels move with it, so a local minimum is reported the same way.

Notation

  1. A local extremum is a local maximum or minimum. Its location is the input \(c\) where it occurs.

    \[\cy{x = c}\]

  2. The value is the output \(f(c)\), so substitute \(c\) into \(f\), not into \(f'\). The lower graph shows that \(f'(c)\) is 0, which is not the height of the top.

    \[\cy{f(c)}\]

  3. The point is the input and the output written together.

    \[\left(c,\ f(c)\right)\]

  4. Match the answer to the question. "Where" and "at what x" ask for c. "What is the value" asks for f(c). "What are the coordinates" asks for the point.

  5. Report an answer only after the sign of \(f'\) changes at \(c\). The curve is green where \(f'\) is positive and red where it is negative, so the color changes at \(c\).

When it fails

At the right end of the slider, the curve still passes through \((2, 3)\) with a horizontal tangent. The location \(x = 2\) and the height 3 both exist.

But on the lower graph \(f'\) stays above the axis on both sides of 2. Its sign does not change, so there is no local maximum to report.

Example 1: location, value and point

Find the location, the value and the point of the local minimum of \(f(x) = e^x - 3x\).

  1. Find the candidate, the input where \(f'\) is 0.

    \[f'(x) = e^x - 3 = 0 \;\Rightarrow\; x = \ln 3\]

  2. Test one point on each side. On the sign chart under the graphs, green is positive and red is negative. The signs are negative, then positive, so \(f\) has a local minimum at \(\ln 3\).

    \[\begin{gathered} f'(0) = \cr{-2} < 0 \\ f'(2) = \cg{e^2 - 3} > 0 \end{gathered}\]

  3. The location is the input.

    \[x = \cy{\ln 3}\]

  4. The value is \(f\) at that input, where \(e^{\ln 3} = 3\). As a check, the value is about \(-0.30\), the dashed line just below the x-axis.

    \[f(\ln 3) = e^{\ln 3} - 3\ln 3 = 3 - 3\ln 3\]

  5. The point has both numbers.

    \[\left(\ln 3,\ 3 - 3\ln 3\right)\]

Example 2: the value, not the input

Find the local maximum value of \(f(x) = -x^3 + 3x^2 + 9x\), and the point where it occurs.

  1. Factor the derivative. It is 0 at \(x = -1\) and \(x = 3\), the two candidates.

    \[\begin{aligned} f'(x) &= -3x^2 + 6x + 9 \\ &= -3(x - 3)(x + 1) \end{aligned}\]

  2. Test \(x = -2\), \(x = 0\) and \(x = 4\). The signs are negative, positive and negative, so \(f\) has a local minimum at \(-1\) and a local maximum at 3.

    \[\begin{gathered} f'(-2) = \cr{-15} \\ f'(0) = \cg{9} \\ f'(4) = \cr{-15} \end{gathered}\]

  3. A tempting wrong move: The local maximum is at \(x = 3\), so the local maximum value is 3.

    \[\text{local maximum value} = 3\]

    The red mark at height 3 on the vertical axis is far below the top, so 3 is not its height. The number 3 is the input, a place on the horizontal axis.

  4. The value is the height of the top, \(f(3)\). Substitute 3 into \(f\).

    \[f(3) = -27 + 27 + 27 = \cy{27}\]

  5. The local maximum value is 27. It occurs at the point \((3, 27)\).

Practice

Find the location and the value of the local maximum of \(g(x) = \sin x + \cos x\) for \(0 < x < 2\pi\). Write the point.

Substitute the location into \(g\), not into \(g'\). The value is the height of the top on the graph.

Solution

  1. Set \(g'\) equal to 0, so \(\cos x = \sin x\). In the interval, this happens at \(x = \frac{\pi}{4}\) and \(x = \frac{5\pi}{4}\), the two candidates.

    \[g'(x) = \cos x - \sin x\]

  2. Test one point in each stretch. The signs are positive, negative and positive, so the local maximum is at \(\frac{\pi}{4}\).

    \[\begin{gathered} g'\!\left(\tfrac{\pi}{6}\right) = \cg{\tfrac{\sqrt3 - 1}{2}} \\ g'\!\left(\tfrac{\pi}{2}\right) = \cr{-1} \\ g'\!\left(\tfrac{3\pi}{2}\right) = \cg{1} \end{gathered}\]

  3. The location is \(x = \frac{\pi}{4}\). The value is \(g\) at that input.

    \[g\!\left(\tfrac{\pi}{4}\right) = \tfrac{\sqrt2}{2} + \tfrac{\sqrt2}{2} = \sqrt2\]

  4. The point is \(\left(\frac{\pi}{4},\ \sqrt2\right)\).

Questions

The graph shows a function m. It has a local maximum at the top and a local minimum at the bottom.

  1. At which input does \(m\) have its local minimum?

    • 2
    • −2
    • −1
    Answer

    2

  2. Which point is the local maximum of \(m\)?

    • (−1, 5)
    • (5, −1)
    • (2, −2)
    Answer

    (−1, 5)

  3. What is the local minimum value of \(h(x) = \dfrac{x^3}{3} - 4x\)?

    • \(-\dfrac{16}{3}\)
    • 2
    • 0
    • \(\dfrac{16}{3}\)
    Answer

    \(-\dfrac{16}{3}\)

Summary

Once the sign of \(f'\) confirms a local maximum or minimum at \(c\), the wording of the question decides which answer you give.

The input \(c\) is a position on the horizontal axis, and the value \(f(c)\), the height there, comes from substituting \(c\) into \(f\).

You answer "where" with the input \(c\), "what value" with \(f(c)\), and "what point" with \((c, f(c))\).

This is one of four skills in topic 5.4 of Which & Why: AP Calculus.