Peaks & Valleys

Where the slope is zero, something interesting is happening.

10 minUpdated 21 October 2026maximumminimumsecond derivativestationary point

Lesson at a glanceShow the key points

At a maximum or minimum of a smooth curve, dy/dx = 0.
d²y/dx² < 0 at a maximum and d²y/dx² > 0 at a minimum.
If the physical situation already makes the type obvious, the second-derivative test can be skipped.
A ball thrown up at speed u peaks at t = u/g, reaching a height u²/2g.

The highest point of a ball’s flight, the cheapest shape for a can, the angle that throws farthest: optimisation problems all come down to finding where a slope is zero.

Flat tops and flat bottoms

Picture any smooth curve with a hump and a dip. At the very top of the hump (a maximum) and the very bottom of the dip (a minimum), the tangent line lies perfectly flat. A flat tangent has zero slope, and slope is the derivative. So at a maximum or minimum,

Condition for a peak or valley

\frac{dy}{dx} = 0

Interactive

Hunting for peaks and valleys

Reshape y = x³ + bx² + cx. Green stretches rise, coral stretches fall. Wherever the colour flips, the slope passes through zero and you get a peak or a valley.

maximum at x = -0.26y″ = -4.58

minimum at x = 1.26y″ = 4.58

Coefficient

b

-1.5

Coefficient

c

-1.0

Peak or valley? Ask the second derivative

A zero slope alone does not say which one you have found. Look at how the slope behaves as you pass through the point:

Approaching a maximum the slope is positive, at the top it is zero, and afterwards it is negative. The slope is decreasing.
Through a minimum the slope goes from negative to zero to positive. The slope is increasing.

“How fast the slope changes” is the derivative of the derivative, written $\dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right) = \dfrac{d^2y}{dx^2}$ . That gives the test:

\text{maximum: } \frac{dy}{dx} = 0,\ \frac{d^2y}{dx^2} < 0 \qquad\qquad \text{minimum: } \frac{dy}{dx} = 0,\ \frac{d^2y}{dx^2} > 0

Interactive

Ball thrown straight up

The height h(t) = ut − ½gt² peaks exactly where its slope dh/dt = u − gt hits zero.

t

0.00s

h

0.00m

dh/dt

15.00m/s

top at t = u/g1.53s

max height11.48m

Launch speed

u

15.0m/s

Time0.00s

Worked example

When does the ball peak?

A ball is thrown straight up at $u = 15\text{ m/s}$ . Its height after time $t$ is $h = ut - \tfrac12 g t^2$ with $g = 9.8\text{ m/s}^2$ . When is it highest, and how high does it go?

Worked example

The roomiest fence

You have 20 m of fencing to enclose a rectangular garden. Which dimensions give the largest area?

What to remember

1At a maximum or minimum, $dy/dx = 0$ .
2 $d^2y/dx^2 < 0$ → maximum; $d^2y/dx^2 > 0$ → minimum.
3If the physical situation makes it obvious, you can skip the second test.

Check yourself

For y = x² − 6x + 11, where is the minimum?

At the top of its flight, a ball thrown straight up has…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

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PreviousDerivatives: Measuring Change Next Integration: Adding Up Slivers

Peaks & Valleys

Where the slope is zero, something interesting is happening.

10 minUpdated 21 October 2026maximumminimumsecond derivativestationary point

Lesson at a glanceShow the key points

At a maximum or minimum of a smooth curve, dy/dx = 0.
d²y/dx² < 0 at a maximum and d²y/dx² > 0 at a minimum.
If the physical situation already makes the type obvious, the second-derivative test can be skipped.
A ball thrown up at speed u peaks at t = u/g, reaching a height u²/2g.

Flat tops and flat bottoms

Condition for a peak or valley

\frac{dy}{dx} = 0

Interactive

Hunting for peaks and valleys

Reshape y = x³ + bx² + cx. Green stretches rise, coral stretches fall. Wherever the colour flips, the slope passes through zero and you get a peak or a valley.

maximum at x = -0.26y″ = -4.58

minimum at x = 1.26y″ = 4.58

Coefficient

b

-1.5

Coefficient

c

-1.0

Peak or valley? Ask the second derivative

A zero slope alone does not say which one you have found. Look at how the slope behaves as you pass through the point:

Approaching a maximum the slope is positive, at the top it is zero, and afterwards it is negative. The slope is decreasing.

Through a minimum the slope goes from negative to zero to positive. The slope is increasing.

“How fast the slope changes” is the derivative of the derivative, written

\dfrac{d}{dx}\!\left(\dfrac{dy}{dx}\right) = \dfrac{d^2y}{dx^2}

. That gives the test:

\text{maximum: } \frac{dy}{dx} = 0,\ \frac{d^2y}{dx^2} < 0 \qquad\qquad \text{minimum: } \frac{dy}{dx} = 0,\ \frac{d^2y}{dx^2} > 0

Interactive

Ball thrown straight up

The height h(t) = ut − ½gt² peaks exactly where its slope dh/dt = u − gt hits zero.

t

0.00s

h

0.00m

dh/dt

15.00m/s

top at t = u/g1.53s

max height11.48m

Launch speed

u

15.0m/s

Time0.00s

Worked example

When does the ball peak?

A ball is thrown straight up at $u = 15\text{ m/s}$ . Its height after time $t$ is $h = ut - \tfrac12 g t^2$ with $g = 9.8\text{ m/s}^2$ . When is it highest, and how high does it go?

Worked example

The roomiest fence

You have 20 m of fencing to enclose a rectangular garden. Which dimensions give the largest area?

What to remember

1At a maximum or minimum, $dy/dx = 0$ .
2 $d^2y/dx^2 < 0$ → maximum; $d^2y/dx^2 > 0$ → minimum.
3If the physical situation makes it obvious, you can skip the second test.

Check yourself

For y = x² − 6x + 11, where is the minimum?

At the top of its flight, a ball thrown straight up has…

Flat tops and flat bottoms#

Peak or valley? Ask the second derivative#

Check yourself

Keep exploring

Flat tops and flat bottoms#

Peak or valley? Ask the second derivative#

Check yourself

Keep exploring

Flat tops and flat bottoms

Peak or valley? Ask the second derivative

Flat tops and flat bottoms

Peak or valley? Ask the second derivative