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 $f^{\prime }\left(x\right)=x^2-x-2=\left(x+1\right)\left(x-2\right)$f'(x) = x² − x − 2 = (x + 1)(x − 2). 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 $x=\frac{1}{2}$x = 1/2, halfway between its roots. To the left of $x=\frac{1}{2}$x = 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 $x=\frac{1}{2}$x = 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 $f\left(x\right)=\frac{x^3}{3}-\frac{x^2}{2}-2x$f(x) = x³/3 − x²/2 − 2x, taking the constant as 0. Then $f\left(-1\right)=-\frac{1}{3}-\frac{1}{2}+2=\frac{7}{6}$f(−1) = −1/3 − 1/2 + 2 = 7/6, $f\left(2\right)=\frac{8}{3}-2-4=-\frac{10}{3}$f(2) = 8/3 − 2 − 4 = −10/3, and $f\left(\frac{1}{2}\right)=\frac{1}{24}-\frac{1}{8}-1=-\frac{13}{12}$f(1/2) = 1/24 − 1/8 − 1 = −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 $f\left(x\right)=\frac{x^3}{3}-\frac{x^2}{2}-2x$f(x) = x³/3 − x²/2 − 2x, with its maximum at x = −1, its minimum at x = 2 and its inflection at $x=\frac{1}{2}$x = 1/2. The faint curve is $f^{\prime }\left(x\right)=x^2-x-2$f'(x) = x² − x − 2, which crosses the axis at −1 and 2 and is lowest at $\frac{1}{2}$1/2.

When f' only touches the axis

Now take $f^{\prime }\left(x\right)={\left(x-1\right)}^2$f'(x) = (x − 1)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 $f\left(x\right)=\frac{{\left(x-1\right)}^3}{3}$f(x) = (x − 1)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'.