Wemazi

Topic 5.9

Choose conclusions supported by the available information

Builds on connecting the zeros of f, f′ and f″ with features. 6 minutes.

The same slope, many curves

The top graph shows \(f'\), the slope of an unknown function \(f\). The lower graph shows curves that could be \(f\), faint copies of one curve shifted up or down.

Move the slider. The teal curve slides up and down, and the short orange marks show the slope that \(f'\) gives at four values of \(x\). The marks fit the curve at every height.

The graph of \(f'\) gives the shape of \(f\), not its height. At every height, \(f\) rises where \(f'\) is positive, falls where \(f'\) is negative, and is flat at \(x = -2\) and \(x = 2\).

Any slope graph

Take any graph of f′. Move the teal curve with C, and move x. The orange bracket shows the vertical gap between the teal curve and the pale one, which is the same at every x.

The gap stays the same because both curves have the same slope everywhere. This means that two functions with the same derivative differ by a constant.

Press "Give one value of f". The known value \(f(a) = b\) marks one point on the lower graph, and only one copy passes through it.

Notation

  1. If \(f\) and \(g\) have the same derivative on an interval, they differ by a constant there.

    \[g(x) = f(x) + \cq{C}\]

  2. The derivative \(f'\) fixes the shape of \(f\): where it rises, falls and bends. It does not fix the height.

  3. One known value \(f(a) = b\) fixes the constant, and so the height.

    \[C = b - F(a)\]

    Here \(F\) is the copy with \(C = 0\), and the known value \(f(0) = 2\) gives \(C = 2\).

  4. A sign of \(f''\) gives a bound, not a value. If \(f'' < 0\) on an interval, the curve lies below its tangent line there.

    \[f(x) < f(a) + f'(a)(x - a)\]

    Here \(f'\) falls for \(x < 0\), so \(f'' < 0\) and the teal curve lies below its orange tangent at \(x = -1.5\). If \(f'' > 0\), the reverse holds.

When it fails

The rule needs one interval. For \(f'(x) = \frac{1}{x^2}\), the graph breaks at \(x = 0\), so the two sides are two intervals.

On each side, \(f(x) = -\frac1x + C\) with its own constant. The value \(f(1) = 0\) fixes the right piece, with \(C_2 = 1\), and tells nothing about \(f(-1)\).

Example 1: find f(2)

A function \(f\) has \(f'(x) = 6x^2 - 2\) for all \(x\), and \(f(1) = 5\). One function with this derivative is \(F(x) = 2x^3 - 2x\). Find \(f(2)\).

  1. Every function with this derivative differs from \(F\) by a constant, so \(f\) is \(F\) plus a constant \(C\).

    \[f(x) = 2x^3 - 2x + C\]

  2. The teal curve must pass through the point \((1, 5)\), and this fixes the constant.

    \[2 - 2 + C = 5, \qquad C = 5\]

  3. With \(C\) known, \(f(2)\) follows.

    \[f(2) = 16 - 4 + 5 = 17\]

    The difference \(f(2) - f(1) = 12\) is the same for every \(C\), because the constant cancels.

Example 2: what is f(4)?

A function has \(f(1) = 10\) and \(f'(1) = 3\), and \(f''(x) < 0\) for \(x \geq 1\). What can you say about \(f(4)\)?

  1. The tangent line at \(x = 1\) is the line through \((1, 10)\) with slope 3. At \(x = 4\), its height is 19.

    \[10 + 3(4 - 1) = 19\]

  2. A tempting wrong move: The slope at \(x = 1\) is 3, so \(f\) rises 9 over the next 3 units and \(f(4) = 19\).

    \[f(4) = 10 + 9 = 19\]

    The slope 3 holds only at \(x = 1\). Because \(f'' < 0\), the slope falls after that, so the curve lies below its tangent line. The red 19 at \(x = 4\) is above the curve.

  3. The data support only a bound, \(f(4) < 19\). The value cannot be found, because \(f\) may bend down slightly or strongly.

    \[f(4) < 19\]

Practice

Let \(g(x) = x^3 + 6x^2 + 9x + C\). Find the value of \(C\) that puts the local minimum of \(g\) on the \(x\)-axis.

Slide \(C\) until the bottom of \(g\) touches the \(x\)-axis. The \(x\)-value of the bottom does not change.

Solution

  1. The constant \(C\) drops out of \(g'\), so \(g'\) is the same for every \(C\). It changes from negative to positive at \(-1\), so the local minimum is at \(x = -1\).

    \[\begin{aligned} g'(x) &= 3x^2 + 12x + 9 \\ &= 3(x + 3)(x + 1) \end{aligned}\]

  2. The minimum value is \(g(-1) = -1 + 6 - 9 + C = C - 4\).

  3. The minimum is on the \(x\)-axis when \(C - 4 = 0\), so \(C = 4\).

Questions

The top graph shows \(f'\). The lower graph shows copies of one curve that have this slope.

  1. Which statement does the graph of \(f'\) support?

    • \(f\) has a local maximum at \(x = \sqrt{3}\)
    • \(f(0) = 3\)
    • \(f(\sqrt{3}) = 0\)
    • \(f(-\sqrt{3}) > f(\sqrt{3})\)
    Answer

    \(f\) has a local maximum at \(x = \sqrt{3}\)

  2. The functions \(f\) and \(g\) both have this \(f'\). Also \(f(0) = 1\) and \(g(0) = 5\). What is \(g(2) - f(2)\)?

    • 4
    • 0
    • 2
    • It cannot be found
    Answer

    4

  3. A function has \(f(1) = 8\) and \(f'(1) = -2\), and \(f''(x) > 0\) for \(x \geq 1\). What is supported about \(f(3)\)?

    • \(f(3) > 4\)
    • \(f(3) = 4\)
    • \(f(3) < 4\)
    • Nothing can be said
    Answer

    \(f(3) > 4\)

Summary

You state only what the data support: the derivatives give the shape of \(f\), and one known value gives its height.

This works because two functions with the same derivative differ by a constant, so their graphs are vertical shifts of each other.

You use it when a problem asks what must be true of \(f\), or whether a value of \(f\) can be found.

This is one of five skills in topic 5.9 of Which & Why: AP Calculus.