JEE Maths›Application of Derivatives›Visual Solution
Visual SolutionStandard
Question
The number of critical points of f(x)=∣x2−1∣f(x) = |x^2 - 1| is:
Solution Path
Piecewise analysis: f′=0f'=0 at x=0x=0, non-differentiable at x=±1x=\pm 1. Three critical points total.
01Question Setup
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Find the number of critical points of f(x)=∣x2−1∣f(x) = |x^2 - 1|.
Critical points =  ?= \;?
02Piecewise Definition
2/4
f(x)=x2−1f(x) = x^2 - 1 when ∣x∣>1|x| > 1 and f(x)=1−x2f(x) = 1 - x^2 when ∣x∣<1|x| < 1. Non-differentiable at x=±1x = \pm 1.
Two cases: ∣x∣>1|x| > 1 and ∣x∣<1|x| < 1
03Find Critical PointsKEY INSIGHT
3/4
f′(x)=0f'(x) = 0 gives x=0x = 0 (from the 1−x21 - x^2 piece). Also ff is not differentiable at x=±1x = \pm 1. Total: 3 critical points.
x=−1,0,1x = -1, 0, 1: three critical points
04Final Answer
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3 critical points: x=−1x = -1 (corner), x=0x = 0 (local max), x=1x = 1 (corner).
3\boxed{3}
Concepts from this question2 concepts unlocked
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First Derivative Test

EASY

The sign change of f'(x) around a critical point determines whether it is a local maximum or minimum. If f'(x) changes from positive to negative, the point is a local max; if negative to positive, it is a local min.

f′(x) changes +→−  ⟹  local max,f′(x) changes −→+  ⟹  local minf'(x) \text{ changes } + \to - \implies \text{local max}, \quad f'(x) \text{ changes } - \to + \implies \text{local min}

JEE problems frequently ask to classify critical points of polynomial and trigonometric functions, and the first derivative test works even when the second derivative is zero or hard to compute

Local extrema classificationPolynomial optimizationCurve sketching
Practice (14 Qs) →

Monotonicity and Intervals

STANDARD

A function f is strictly increasing on an interval where f'(x) > 0 and strictly decreasing where f'(x) < 0. Finding these intervals requires solving f'(x) = 0 to get critical points, then testing the sign of f'(x) in each sub-interval.

f′(x)>0  ⟹  f is strictly increasing,f′(x)<0  ⟹  f is strictly decreasingf'(x) > 0 \implies f \text{ is strictly increasing}, \quad f'(x) < 0 \implies f \text{ is strictly decreasing}

JEE regularly asks for intervals of increase/decrease, and monotonicity is essential for proving inequalities and determining the number of roots of an equation

Increasing/decreasing intervalsInequality proofsRoot counting problems
Practice (11 Qs) →
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