You already know that a derivative measures slope. That single fact is quietly powerful: if you can read the slope at every point, you can read the shape of the whole curve — where it climbs, where it falls, and exactly where it turns around. Master that, and a whole class of real problems — the most area, the least material, the best price — turns into one clean question: where is the derivative zero?
The sign of the slope tells you the shape
Picture walking left to right along a graph. At each step the tangent line has a slope, and that slope is just \( f'(x) \). Now notice what the slope is telling you about your walk.
If the slope is positive, the tangent tilts uphill — so the curve is rising. If it is negative, the tangent tilts downhill — the curve is falling. We give these everyday words a name:
- Where \( f'(x) > 0 \), the function is increasing.
- Where \( f'(x) < 0 \), the function is decreasing.
So the sign of the derivative is a kind of compass. You do not have to plot a hundred points to know the shape — you just have to ask whether \( f'(x) \) is above or below zero.
In words The derivative does not just give you one number — its sign hands you the behaviour of the function over a whole stretch. Positive means up, negative means down. Read the sign, and you have read the trend.
Where the slope is zero, the curve turns
Something special happens at the exact moment the curve stops rising and starts falling, or the other way round. To switch from a positive slope to a negative one, the slope has to pass through zero. At that instant the tangent line is perfectly flat.
A point where \( f'(x) = 0 \) is called a stationary point — and more generally, any point where the derivative is zero (or fails to exist) is a critical point. These are the only places a smooth curve can have a peak or a valley, so they are exactly the candidates worth checking. The flat tangent is your signal: something interesting happens here.
But a flat tangent alone does not tell you which kind of turn it is. For that you look at the slope on either side.
The first-derivative test
Stand at a critical point and glance just to its left and just to its right.
- If \( f' \) is positive then negative (rising into the point, falling out of it), you are standing on a local maximum — a peak.
- If \( f' \) is negative then positive (falling in, rising out), you are at the bottom of a local minimum — a valley.
That is the whole first-derivative test: a sign change in \( f' \) tells you the type of turning point. It matches your intuition perfectly — a hilltop is where you stop going up and start coming down.
The curve below is \( f(x) = x^3 - 3x \), with its derivative drawn dashed. Slide the point along it: watch the reported slope flip from positive to negative at the peak near \( x = -1 \), and from negative to positive at the valley near \( x = 1 \).
Tip — a sign chart keeps it tidy. Find the critical points, mark them on a number line, then test one easy value of \( f' \) in each gap. The pattern of \( + \) and \( - \) signs reveals every peak and valley at a glance, with no graphing required.
Worked example: classify every turning point
Let us run the test carefully on the very curve in the widget, so the picture and the algebra agree.
- Differentiate: \( f'(x) = 3x^2 - 3 \).
- Set the derivative to zero and factor: \( 3x^2 - 3 = 3(x^2 - 1) = 3(x-1)(x+1) = 0 \). The critical points are \( x = -1 \) and \( x = 1 \).
- Test the sign of \( f' \) on each side. To the left of \( -1 \): \( f'(-2) = 3(4) - 3 = 9 > 0 \). Between them: \( f'(0) = -3 < 0 \). To the right of \( 1 \): \( f'(2) = 3(4) - 3 = 9 > 0 \).
- At \( x = -1 \), \( f' \) goes \( + \to - \): a local maximum. Its height is \( f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2 \).
- At \( x = 1 \), \( f' \) goes \( - \to + \): a local minimum. Its height is \( f(1) = (1)^3 - 3(1) = 1 - 3 = -2 \).
There is a local maximum at \( (-1,\, 2) \) and a local minimum at \( (1,\, -2) \) — exactly the two marked points on the interactive curve.
Optimization: turning a real problem into "where is the slope zero?"
Here is where this idea earns its keep. So many practical questions are really requests for a maximum or a minimum: enclose the largest area with a fixed fence, build a can with the least metal, set the price for the most profit. Each one hides a function — and the best answer sits at one of its turning points.
The recipe is always the same. Write the quantity you care about as a function of one variable, differentiate it, set the derivative to zero, and check the sign change to confirm it is the maximum (or minimum) you wanted.
- The perimeter is fixed: \( 2(\text{width} + \text{height}) = 20 \), so width + height \( = 10 \). Call one side \( x \); the other must be \( 10 - x \).
- Write the area as a function of that single variable: \[ A(x) = x(10 - x) = 10x - x^2. \]
- Differentiate and set it to zero: \( A'(x) = 10 - 2x = 0 \), which gives \( x = 5 \).
- Confirm it is a maximum with the sign change: \( A'(4) = 10 - 8 = 2 > 0 \) and \( A'(6) = 10 - 12 = -2 < 0 \). Since \( A' \) goes \( + \to - \), the area is greatest here.
Both sides equal \( 5 \) — a \( 5 \times 5 \) square — with area \( A(5) = 25 \). Of all rectangles with a given perimeter, the square wins, and calculus tells you why in three short lines.
In words Optimization is not a new technique — it is the first-derivative test pointed at a real question. The hardest step is usually writing the quantity as a function of one variable; once you have that, differentiate, set it to zero, and the answer falls out.
Practice
Try each one yourself, then reveal the full solution.
1. On what intervals is \( f(x) = x^2 - 4x \) increasing or decreasing, and where is its minimum?
Differentiate: \( f'(x) = 2x - 4 \). Set it to zero: \( 2x - 4 = 0 \) gives the critical point \( x = 2 \).
Check the sign on each side. For \( x < 2 \), say \( f'(0) = -4 < 0 \), so the function is decreasing. For \( x > 2 \), say \( f'(3) = 2 > 0 \), so it is increasing. The slope goes \( - \to + \), a local (and here global) minimum.
The value there is \( f(2) = 4 - 8 = -4 \).
Answer: decreasing for \( x < 2 \), increasing for \( x > 2 \); minimum at \( x = 2 \) with value \( -4 \).
2. Find the critical point of \( f(x) = -x^2 + 6x \) and say whether it is a maximum or a minimum.
Differentiate: \( f'(x) = -2x + 6 \). Set it to zero: \( -2x + 6 = 0 \) gives \( x = 3 \).
Test the sign: \( f'(2) = 2 > 0 \) and \( f'(4) = -2 < 0 \). The slope goes \( + \to - \), so this is a maximum.
The value is \( f(3) = -9 + 18 = 9 \).
Answer: critical point at \( x = 3 \); it is a maximum, with value \( 9 \).
3. Two positive numbers add up to 12. What choice makes their product as large as possible?
Let one number be \( x \); the other is \( 12 - x \). Write the product as a function of one variable: \[ P(x) = x(12 - x) = 12x - x^2. \]
Differentiate and set to zero: \( P'(x) = 12 - 2x = 0 \), so \( x = 6 \).
Confirm the maximum: \( P'(5) = 2 > 0 \) and \( P'(7) = -2 < 0 \), a \( + \to - \) sign change. So the product peaks at \( x = 6 \), making the other number \( 12 - 6 = 6 \) as well.
Answer: the numbers are 6 and 6, giving the largest product \( 36 \).