Q1. Find the local maximum/minimum of f(x) = x² − 4x + 3.
f'(x) = 2x − 4; set f'(x) = 0 ⇒ x = 2
f''(x) = 2, which is POSITIVE
So x = 2 gives a local MINIMUM; f(2) = 4 − 8 + 3 = −1
Application of Derivatives — Class 12
Derivatives aren't just formulas — they find the fastest route, the biggest box, the cheapest fence. Master maxima and minima here.
Rate of changedy/dx measures how fast y changes as x changes — e.g. dV/dt for the rate a volume grows.
Increasing / decreasingf is increasing where f'(x) > 0, and decreasing where f'(x) < 0.
Critical pointsPoints where f'(x) = 0 (or is undefined) — candidates for maxima or minima.
Second derivative testAt a critical point: f''(x) < 0 ⇒ local maximum; f''(x) > 0 ⇒ local minimum.
Solved examples
Q2. The side of a cube is increasing at 2 cm/s. Find the rate of increase of its volume when the side is 5 cm.
V = a³, so dV/dt = 3a² · da/dt
dV/dt = 3(5²)(2) = 3 × 25 × 2
dV/dt = 150 cm³/s
Formula sheet
| Quantity | Formula |
|---|---|
| Increasing function | f'(x) > 0 |
| Decreasing function | f'(x) < 0 |
| Local maximum test | f'(x)=0 and f''(x)<0 |
| Local minimum test | f'(x)=0 and f''(x)>0 |
Common mistakes & exam wins
- Always find critical points by setting the FIRST derivative to zero.
- Use the SECOND derivative to classify each critical point as a max or min (negative = max, positive = min).
- For rate-of-change word problems, identify the two related quantities and differentiate the connecting formula with respect to time.
Frequently asked questions
How do you find the maximum or minimum value of a function?
Set the first derivative to zero to find critical points, then use the second derivative test (or a sign check) to classify each as a max or min.
What does it mean for a function to be increasing?
Its derivative f'(x) is positive over that interval — the function's value rises as x increases.
What is the second derivative test?
At a critical point where f'(x) = 0: if f''(x) < 0 it's a local maximum, if f''(x) > 0 it's a local minimum.