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Grade 11-12 (age 16-18)

Applications of Derivatives

Applications of Derivatives

The derivative works like a compass that tells you whether a function's graph is currently climbing uphill or sliding downhill. A positive slope means it is rising, a negative slope means it is falling, and the exact spot where the sign flips is a peak (local max) or a valley (local min). Once you grasp this, a single derivative lets you read off a function's maximum and minimum values, and even the velocity and acceleration of a moving object. Drag the x-position slider and watch how the tangent tips over and how the green (increasing) and red (decreasing) regions trade places.

Increasing vs Decreasing
📈 What the Sign of f' Tells You
①f'(x) > 0 ⇒ f(x) increasing
②f'(x) < 0 ⇒ f(x) decreasing
③f'(x) = 0 ⇒ candidate for local max/min
Extrema & Sign Chart
0
Extremum Test
f'(a) = 0 with sign + → - ⇒ local max, - → + ⇒ local min
Use a sign chart for the derivative
💡 Building a Sign Chart
①Solve f'(x) = 0 (e.g., 3x² − 3 = 0 → x = ±1)
②Determine the sign of f'(x) on each interval
③Sign change marks a local extremum
④No sign change at a critical point ⇒ NOT an extremum (e.g., x³ at x = 0)
Maxima & Minima
Max/Min on a Closed Interval
[a, b] candidates: ①f'(x)=0 points ②endpoints f(a), f(b)
Compare candidate values; max wins, min wins
Second Derivative Test
f'(a)=0, f''(a)<0 ⇒ local max / f'(a)=0, f''(a)>0 ⇒ local min
Decide max/min by the sign of f''
🔑 Max-Min Strategy
①Closed interval: compare extrema + endpoints
②Open interval: extrema alone
③Real-world optimization: set up an objective and differentiate
Velocity & Acceleration
1 s
Position-Velocity-Acceleration
v(t) = s'(t), a(t) = v'(t) = s''(t)
Position → velocity → acceleration
💡 Reading Motion
①v(t) > 0: moving in + direction / v(t) < 0: − direction
②v(t) = 0: turning point (peak)
③Constant a(t) ⇒ uniformly accelerated motion (e.g., free fall)
Wrap-up
Extremum Condition
f'(a) = 0 with sign change ⇒ local max/min
Sign change of f' is what matters
🎯 Exam Points
①Sign chart: solve f'(x) = 0, then check signs on each interval
②Test: f' + → − is max; − → + is min
③Closed interval: compare extrema + endpoints
④Tangent line: y = f(a) + f'(a)(x − a)
⑤Use the sign chart to sketch and solve equations/inequalities
Worked Examples & Past Exam
Example 1
Find the local maximum and minimum of f(x) = x³ − 3x.
1
Find the derivative and solve f'(x) = 0.
f'(x) = 3x2 - 3 = 0 ⟹ x = ±1
2
By the sign of f', x = −1 gives a maximum and x = 1 a minimum.
local max f(-1) = 2, local min f(1) = -2
local max 2, local min −2
Where f'(x) = 0, a sign change +→− gives a max and −→+ gives a min.
Example 2
A point P starts at the origin with position s(t) = t² − 4t. Find the velocity v(1) at t = 1.
1
Velocity is the derivative of position. v(t) = s'(t).
v(t) = s'(t) = 2t - 4
2
Substitute t = 1.
v(1) = 2(1) - 4 = -2
v(1) = −2
Differentiating position once gives velocity; its sign shows the direction of motion.
2022 KICE mock exam Math type, adapted
What is the slope of the tangent to y = x² + 1 at the point (1, 2)?
2
1
3
4
0
① 2
1
The tangent slope is the derivative at that point. y' = 2x.
y' = (x2 + 1)' = 2x
2
Substitute x = 1.
y'(1) = 2(1) = 2
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Indefinite Integral
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