Reading derivatives from graphs, tangent lines, and difference quotients

Reading the steepness and direction of the tangent line at each marked \(x\)-value:

(a) At \(x=-3\) the curve is falling steeply, so \(f'(-3)\) is a large negative number, roughly \(f'(-3) \approx -3\text{.}\)

(b) At \(x=-1\) the curve is on the shallow flat shelf, so the tangent line is nearly horizontal: \(f'(-1) \approx 0\text{.}\)

(c) At \(x=1\) the curve is still gently decreasing on its way down to the minimum, so \(f'(1) \approx -1\) (negative, but less steep than at \(x=-3\)).

(d) At \(x=3\) the curve is climbing very steeply out of the minimum, so \(f'(3)\) is a large positive number, roughly \(f'(3) \approx 5\text{.}\)

(Exact values will vary slightly from student to student — what matters is the sign and relative size of each estimate.)

Consider the graph of \(f(x)\) shown. For each part, give the relationship between the two quantities: \(\lt\text{,}\) \(=\text{,}\) \(\gt\text{,}\) or unclear.

1. \(f(6)\) ____ \(f(8)\)
2. \(f(6) - f(4)\) ____ \(f(4) - f(2)\)
3. \(\dfrac{f(4)-f(2)}{4-2}\) ____ \(\dfrac{f(6)-f(2)}{6-2}\)
4. \(f'(2)\) ____ \(f'(8)\)

\(f\) is increasing and convex (bends upward), so its slope is increasing and its rate of growth speeds up to the right.

(a) Since \(f\) is increasing, \(f(6) \lt f(8)\text{.}\)

(b) Because \(f\) is convex, equal-width intervals farther to the right show a bigger change, so \(f(6) - f(4) \gt f(4) - f(2)\text{.}\)

(c) The secant slope over a longer interval starting at the same point picks up the steeper right-hand part of the curve, so \(\dfrac{f(4)-f(2)}{4-2} \lt \dfrac{f(6)-f(2)}{6-2}\text{.}\)

(d) Since the curve gets steeper as \(x\) increases, \(f'(2) \lt f'(8)\text{.}\)

The graph shown is \(f'(x)\text{,}\) not \(f(x)\) itself — so parts (a)/(b) and parts (c)/(d) need two different kinds of reasoning.

(a), (b) Notice the graph of \(f'(x)\) never dips below the axis — it only touches \(0\) briefly at \(x_6\) and is positive everywhere else. Since \(f'(x) \ge 0\) throughout, \(f\) is non-decreasing over the whole picture. For a non-decreasing function, the smallest labeled \(x\)-value gives the smallest output and the largest labeled \(x\)-value gives the largest output. So \(f(x)\) is least at \(x = x_1\) (the leftmost label) and greatest at \(x = x_7\) (the rightmost label).

(c), (d) Now we read the graph directly, since it is \(f'(x)\text{.}\) The lowest point on the graph among the labeled values is where it dips to touch \(0\) at \(x_6\text{,}\) so \(f'(x)\) is least at \(x = x_6\). The highest point is the peak of the hump at \(x_3\text{,}\) so \(f'(x)\) is greatest at \(x = x_3\).

Let \(g(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}\text{,}\) where \(f(x)\) is the piecewise linear function shown. Find:

1. \(\displaystyle g(0)\)
2. \(\displaystyle g(3)\)
3. \(\displaystyle g(5)\)
4. \(\displaystyle g(8)\)

\(g(x) = f'(x)\text{.}\) Since \(f\) is piecewise linear, \(f'\) is constant on each linear piece and undefined at the corners.

(a) \(x=0\) lies on the segment from \((-3,-3)\) to \((2,1)\text{,}\) with slope \(\dfrac{1-(-3)}{2-(-3)} = \dfrac{4}{5}\text{.}\) So \(g(0) = \dfrac{4}{5}\text{.}\)

(b) \(x=3\) lies on the flat segment from \((2,1)\) to \((4,1)\text{,}\) which has slope \(0\text{.}\) So \(g(3) = 0\text{.}\)

(c) \(x=5\) is the corner where the rising segment (slope \(3\)) meets the falling segment (slope \(\approx -3\)). Since the left- and right-hand slopes disagree, \(f'(5)\) does not exist, so \(g(5)\) does not exist.

(d) \(x=8\) lies on the falling segment from \((5,4)\) to about \((8.3,-6)\text{,}\) with slope \(\dfrac{-6-4}{8.3-5} \approx -3\text{.}\) So \(g(8) \approx -3\text{.}\)

Suppose that \(f(x)\) is a function with \(f(90) = 58\) and \(f'(90) = 3\text{.}\)

Estimate \(f(93.5)\text{.}\)

Use the tangent line (linear) approximation at \(x=90\text{:}\)

\begin{equation*} f(93.5) \approx f(90) + f'(90)\,(93.5 - 90) \end{equation*}

\begin{equation*} f(93.5) \approx 58 + 3(3.5) = 58 + 10.5 = 68.5 \end{equation*}

Let \(f(x) = 2 + 6x^2\text{.}\) If \(h \neq 0\text{,}\) then the difference quotient can be simplified as

where \(A\text{,}\) \(B\text{,}\) and \(C\) are constants.

1. Find the constants \(A\text{,}\) \(B\text{,}\) and \(C\text{.}\)
2. Use the simplified expression to find \(f'(x) = \displaystyle\lim_{h \to 0} \dfrac{f(x+h) - f(x)}{h}\text{.}\)

\begin{equation*} f(x+h) = 2 + 6(x+h)^2 = 2 + 6x^2 + 12xh + 6h^2 \end{equation*}

\begin{equation*} f(x+h) - f(x) = 12xh + 6h^2 \end{equation*}

\begin{equation*} \dfrac{f(x+h)-f(x)}{h} = 12x + 6h = 6h + 12x + 0 \end{equation*}

(a) Matching to \(Ah + Bx + C\text{:}\) \(A = 6\text{,}\) \(B = 12\text{,}\) \(C = 0\text{.}\)

(b) Taking \(h \to 0\text{:}\)

\begin{equation*} f'(x) = 12x \end{equation*}

Given:

• \(f'(x) \gt 0\) if \(x \lt -3\)
• \(f'(x) \lt 0\) if \(-3 \lt x \lt 3\)
• \(f'(x) \gt 0\) if \(x \gt 3\)
• \(\displaystyle |f(x)| \lt 6\)

Use the information provided to sketch a possible graph of \(f(x)\text{.}\)

The sign of \(f'\) tells us where \(f\) is increasing or decreasing:

\(f\) is increasing on \((-\infty, -3)\text{,}\) reaches a local maximum at \(x=-3\), then is decreasing on \((-3, 3)\text{,}\) reaches a local minimum at \(x=3\), then is increasing again on \((3, \infty)\text{.}\)

The condition \(|f(x)| \lt 6\) means the whole graph must stay strictly between the horizontal lines \(y=6\) and \(y=-6\) — so a valid sketch rises toward (but never reaches) some value below \(6\) at \(x=-3\text{,}\) dips toward (but never reaches) some value above \(-6\) at \(x=3\text{,}\) and continues to rise afterward while staying under \(y=6\text{.}\)

Suppose that \(f(x) = 2x^2 - 8x\text{.}\)

1. Find the equation of the line tangent to \(f(x)\) at \(x = 2\text{.}\) \(y =\) ____

The slope of the tangent line at \(x=2\) is \(f'(2)\text{.}\) Differentiating,

\begin{equation*} f'(x) = 4x - 8 \end{equation*}

\begin{equation*} f'(2) = 4(2) - 8 = 0 \end{equation*}

The slope is \(0\text{,}\) so the tangent line is horizontal. The point of tangency is \(f(2) = 2(4) - 8(2) = 8 - 16 = -8\text{,}\) so the point is \((2, -8)\text{.}\) A horizontal line through \((2,-8)\) is simply:

\begin{equation*} y = -8 \end{equation*}

The displacement (in meters) of a particle moving in a straight line is given by

where \(t\) is measured in seconds. Find the instantaneous velocity when \(t = 4\text{.}\)

The instantaneous velocity is \(s'(t)\text{.}\) Differentiating,

\begin{equation*} s'(t) = 2t - 5 \end{equation*}

So at \(t=4\text{,}\)

\begin{equation*} s'(4) = 2(4) - 5 = 3 \text{ m/s} \end{equation*}