From the Derivative Back to the Function

Reading a graph of the gradient

Sometimes the graph you are given is y = f′(x), and the question is about f. Each height on that graph is a gradient: the height of f′ at x is how steeply f is rising or falling at x.

So the graph of f′ never tells you how high f is, only which way it is going and how fast. Read it for three things: where it is zero, where it is above or below the x-axis, and where it turns.

Zeros and sign

Take $f'\left(x\right) = x^{2} - x - 2 = \left(x + 1\right)\left(x - 2\right)$. It is zero at x = −1 and x = 2, so f has a horizontal tangent at both: they are its stationary points.

Between the zeros, at x = 0, f′(0) = −2 is negative, so f is decreasing on −1 < x < 2. Outside them, at x = −2 or x = 3, f′ = 4, positive, so f is increasing for x < −1 and for x > 2.

Maximum or minimum

At x = −1, f′ changes from positive to negative: f rises up to x = −1 and falls after it, so x = −1 is a maximum of f.

At x = 2, f′ changes from negative to positive: f falls down to x = 2 and rises after it, so x = 2 is a minimum. This is the first derivative test, read straight off a picture.

The turning point of f′

f' is lowest at $x = \frac{1}{2}$, halfway between its roots. To the left of $x = \frac{1}{2}$, f' is falling, so f'' < 0 and f is concave down; to the right f' is rising, so f'' > 0 and f is concave up.

So f changes concavity at $x = \frac{1}{2}$: a point of inflection. A turning point of f' is an inflection of f, in the same way that a zero of f' is a stationary point of f.

Drawing f

Here f can be found exactly: integrating gives $f\left(x\right) = \frac{x^{3}}{3} - \frac{x^{2}}{2} - 2x$, taking the constant as 0. Then $f\left(-1\right) = - \frac{1}{3} - \frac{1}{2} + 2 = \frac{7}{6}$, $f\left(2\right) = \frac{8}{3} - 2 - 4 = - \frac{10}{3}$, and $f\left(\frac{1}{2}\right) = \frac{1}{24} - \frac{1}{8} - 1 = - \frac{13}{12}$.

A different constant would slide the whole graph up or down. The graph of f′ fixes its shape, not its height.

The gold curve is $f\left(x\right) = \frac{x^{3}}{3} - \frac{x^{2}}{2} - 2x$, with its maximum at x = −1, its minimum at x = 2 and its inflection at $x = \frac{1}{2}$. The faint curve is $f'\left(x\right) = x^{2} - x - 2$, which crosses the axis at −1 and 2 and is lowest at $\frac{1}{2}$.

When f′ only touches the axis

Now take $f'\left(x\right) = \left(x - 1\right)^{2}$. It is zero at x = 1, so f is flat there, but it touches the axis without crossing it: f' is positive on both sides.

So f rises, flattens for an instant at x = 1, and rises again. That stationary point is neither a maximum nor a minimum; since x = 1 is also the lowest point of f', it is a stationary point of inflection. Here $f\left(x\right) = \frac{\left(x - 1\right)^{3}}{3}$, plus any constant.

The usual slips

Reading the sign of f′ as the sign of f: below the axis on a graph of f′ means f is decreasing, not that f is negative.

Reading "f′ is falling" as "f is decreasing": a falling f′ only means f is bending downward. f decreases where f′ is negative, wherever f′ is heading.

Calling every zero of f′ an inflection: a zero of f′ is a stationary point of f. The inflections of f sit at the turning points of f′.