Grade 11-12 (age 16-18)

high school Applications of Derivatives

The sign of the derivative tells you whether a graph is going uphill or downhill at the moment. A positive slope means rising and a negative slope means falling, so the point where the sign changes is a peak, a local maximum, or a valley, a local minimum. That one principle lets you read off the largest and smallest values of a function, and even the velocity and acceleration of a moving object, from differentiation. In the first picture, moving the x position matches the slope of the tangent with the green (increasing) and red (decreasing) bands; in the second, position, velocity and acceleration are read together at each time t.

Telling increasing from decreasing

📈 What the sign of the derivative tells you

①Where f'(x) > 0, f(x) is increasing (uphill)

②Where f'(x) < 0, f(x) is decreasing (downhill)

③Points with f'(x) = 0 are candidates for a local maximum or minimum

Local extrema and the sign chart

x position0

Identifying local extrema

f'(a) = 0 and f' changes sign + → −: local maximum; − → +: local minimum

Draw a sign chart to check whether the derivative changes sign

💡 Building a sign chart

①Solve f'(x) = 0 (for example, 3x² − 3 = 0 gives x = ±1)

②Fix the sign of f'(x) on each interval between the roots

③Where the sign changes, decide whether it is a maximum or a minimum

④f'(x) = 0 without a sign change is not an extremum; for example, y = x³ has no extremum at x = 0

Maximum and minimum values

Maximum and minimum on a closed interval

Candidates on [a, b]: ① points with f'(x) = 0 ② the endpoints f(a), f(b)

The largest function value among the candidates is the maximum, and the smallest is the minimum

The second-derivative test

f'(a)=0, f''(a)<0 ⇒ local max / f'(a)=0, f''(a)>0 ⇒ local min

The sign of the second derivative separates maxima from minima; inflection points come with second derivatives in calculus

🔑 Finding maxima and minima

①On a closed interval, compare the values at local extrema, at the endpoints and at points where f is not differentiable

②On an open interval there may be no maximum or minimum

③For real-life optimization, write the quantity to optimize as a function and differentiate to find its extrema

Velocity & Acceleration

Time t1 s

Position, velocity and acceleration

v(t) = s'(t), a(t) = v'(t) = s''(t)

Differentiate position once for velocity and twice for acceleration

💡 Reading the motion

①If v(t) > 0 the object moves in the positive direction; if v(t) < 0, in the negative direction

②In the picture v = 0 at t = 2, and that is when the position s is greatest

③If v(t) = 0 but its sign does not change, the direction stays the same

④If a(t) is constant, the motion has constant acceleration

Summary

Condition for an extremum

f'(a) = 0 with a change of sign → local maximum or minimum

What matters is whether the derivative changes sign

🎯 Exam Points

①Sign chart: solve f'(x) = 0 and check the sign on each interval

②f' changing from + to − gives a local maximum; from − to +, a local minimum

③On a closed interval, compare extrema and endpoints for the maximum and minimum

④The tangent line is y = f'(a)(x − a) + f(a)

⑤For equations and inequalities, sketch the shape of the graph from a sign chart first

Worked Examples & Past Exam

Example 1

Find the local maximum and minimum of f(x) = x³ − 3x.

Differentiate and find the roots of f'(x) = 0.

f'(x) = 3x² - 3 = 0 ⟹ x = ±1

f' > 0 for x < −1, f' < 0 for −1 < x < 1 and f' > 0 for x > 1, so there is a local maximum at x = −1 and a local minimum at x = 1.

local max f(-1) = 2, local min f(1) = -2

▸ local max 2, local min −2

The green and red bands in the first picture are exactly this sign chart. Where the sign goes from + to − there is a peak; where it goes from − to + there is a valley.

Example 2

A point P starts at the origin with position s(t) = t² − 4t. Find the velocity v(1) at t = 1.

Velocity is the derivative of position with respect to time.

v(t) = s'(t) = 2t - 4

Put t = 1.

v(1) = 2(1) - 4 = -2

▸ v(1) = −2

v(1) is negative, so at t = 1 the point P is moving in the negative direction. At t = 2, v = 0, and after that the direction reverses.

2022 KICE mock exam Math type, adapted

What is the slope of the tangent to y = x² + 1 at the point (1, 2)?

①2

②1

③3

④4

⑤0

▸ ① 2

The slope of the tangent is the derivative at that point, and y' = 2x.

y' = (x² + 1)' = 2x

Put x = 1.

y'(1) = 2(1) = 2

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Last updated 2026-10-17

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