MCV4U Graphs of f, f′ and f″: Notes and Practice

Often you have a graph and no equation: a velocity–time graph from a motion sensor, a chart of a stock price, a temperature log. Calculus still works, because every feature of a function leaves a fingerprint on its derivatives. A peak in $f$ is a zero crossing in $f^{'}$ . The steepest point of $f$ is a peak or valley of $f^{'}$ and a zero crossing of $f^{''}$ . Once you can see these links, you can sketch $f^{'}$ and $f^{''}$ from the shape of $f$ alone, and you can run the process in reverse.

Sketching f′ and f″ from the graph of f

Where $f$ is increasing, the graph of $f^{'}$ is ABOVE the $x$ -axis; where $f$ is decreasing, $f^{'}$ is BELOW it. (The one exception is a single flat point, like $x^{3}$ at $0$ , where $f^{'}$ touches the axis without crossing.)
At a local maximum or minimum of a smooth $f$ (a horizontal tangent), $f^{'}$ CROSSES the $x$ -axis: from $+$ to $-$ at a maximum, from $-$ to $+$ at a minimum.
Where $f$ is concave up, its slopes are increasing, so $f^{'}$ is INCREASING and $f^{''}$ is above the axis. Where $f$ is concave down, $f^{'}$ is decreasing and $f^{''}$ is below the axis.
At a point of inflection of $f$ , the slope stops increasing and starts decreasing (or the reverse), so $f^{'}$ has a LOCAL MAXIMUM OR MINIMUM and $f^{''}$ crosses the $x$ -axis.
Each derivative of a polynomial has degree one less: a cubic $f$ has a quadratic $f^{'}$ (a parabola) and a linear $f^{''}$ (a line).

Fig 1Three aligned graphs for f(x) = x³ − 6x² + 9x + 1: the maximum and minimum of f line up with the zeros of f′, and the inflection point of f lines up with the minimum of f′ and the zero of f″, shown by dashed vertical guides.

Feature of f	What f′ shows	What f″ shows
Increasing	$f^{'} > 0$ (above the axis), apart from single flat points	no direct information
Decreasing	$f^{'} < 0$ (below the axis), apart from single flat points	no direct information
Local maximum	$f^{'} = 0$ , changing $+$ to $-$	usually $f^{''} < 0$ (concave down)
Local minimum	$f^{'} = 0$ , changing $-$ to $+$	usually $f^{''} > 0$ (concave up)
Concave up	$f^{'}$ increasing	$f^{''} > 0$
Concave down	$f^{'}$ decreasing	$f^{''} < 0$
Point of inflection	$f^{'}$ has a local max or min	$f^{''} = 0$ (or undefined) and changes sign

How the key features of a function appear on the graphs of its first and second derivatives.

Check this against the stacked graphs. For $f (x) = x^{3} - 6 x^{2} + 9 x + 1$ , the derivative $f^{'} (x) = 3 x^{2} - 12 x + 9 = 3 (x - 1) (x - 3)$ crosses zero at $x = 1$ (the maximum $(1, 5)$ ) and $x = 3$ (the minimum $(3, 1)$ ). The parabola $f^{'}$ has its vertex at $x = 2$ , where $f^{'} (2) = - 3$ is the steepest downhill slope of $f$ ; that is the inflection point $(2, 3)$ . And $f^{''} (x) = 6 x - 12$ is a line crossing zero at exactly $x = 2$ .

Translations behave in a predictable way. Shifting $f$ up or down by $k$ does not change any slope, so $f (x) + k$ has exactly the same derivative as $f (x)$ . Shifting $f$ right by $h$ moves every slope right by $h$ too, so the derivative of $f (x - h)$ is $f^{'} (x - h)$ : the graph of $f^{'}$ shifts right by $h$ .

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Pick a function and toggle the graphs of f′ and f″ underneath it; the extrema of f line up with zeros of f′, and the inflection points with the peaks and valleys of f′.

VideoEddie Woo sketches the derivative straight from the shape of a graph by reading where the slope is zero, positive and negative.Watch on YouTube

From f′ back to f

Now reverse the question: you know $f^{'}$ (as a graph, a sign chart or an equation) and you want $f$ . The same links tell you where $f$ rises and falls, where it turns, and where it bends. What they cannot tell you is HOW HIGH the graph sits, because $f^{'}$ only describes slopes.

Fig 2Given f′(x) = 2x − 4, three of the possible functions f(x) = x² − 4x + C, for C = −2, 0 and 2: identical shapes shifted vertically, with parallel tangents of slope 2 at x = 3. The curve through the given point (4, 2), where C = 2, is highlighted.

Why infinitely many graphs are possible

f^{'} (x) = g (x)

, then

f (x) + C

has the same derivative for every constant

C

, since the derivative of a constant is

0

. So knowing

f^{'}

fixes the SHAPE of

f

but not its vertical position: there are infinitely many possible graphs, all vertical translations of one another. One extra fact, such as the value

f (0)

, picks out a single graph.

The curriculum's sample: $g (x) = f^{'} (x) = (x - 1) (x - 3) = x^{2} - 4 x + 3 .$ Then $f^{''} (x) = g^{'} (x) = 2 x - 4$ . Reading features: $f^{'} > 0$ for $x < 1$ and $x > 3$ and $f^{'} < 0$ between, so $f$ has a local maximum at $x = 1$ and a local minimum at $x = 3$ ; $f^{''}$ changes sign at $x = 2$ , so $f$ has an inflection point there. Since $f^{'}$ is a quadratic with positive leading coefficient, $f$ is a cubic with positive leading coefficient. To describe a possible equation, ask which cubic has this derivative: since $\frac{1}{3} x^{3}$ differentiates to $x^{2}$ , a good guess is $\frac{1}{3} x^{3} - 2 x^{2} + 3 x$ , and differentiating it confirms the guess. Adding a constant changes nothing, so every possibility has the form $f (x) = \frac{1}{3} x^{3} - 2 x^{2} + 3 x + C$ . With $f (0) = 0$ , $C = 0$ ; with $f (0) = 2$ , $C = 2$ and the whole graph moves up $2$ units.

Height on f′ means slope on f

When you read a graph of

f^{'}

, its

y

-values are SLOPES of

f

, not heights of

f

. A point where the graph of

f^{'}

is at its highest is where

f

is steepest uphill, not where

f

is highest. The highest point of

f

(nearby) is where

f^{'}

crosses zero going down.

Graphs of f, f′ and f″

Key things to remember

Sketching f′ and f″ from the graph of f

From f′ back to f

Worked examples

Where marks get lost

Flashcards 8

Practice