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Core Calculus Reading Time: 15 min Essential for JEE

The Relationship Between f(x), f'(x), and f''(x) for Differentiability

Understand how functions and their derivatives connect geometrically and analytically for JEE success.

3-4
Questions/JEE Paper
100%
Conceptual Understanding
5
Key Relationships
12min
Avg. Solve Time

Why This Relationship Matters for JEE

Understanding how f(x), f'(x), and f''(x) relate is crucial for solving 3-4 questions in every JEE paper. These relationships help you:

  • Analyze function behavior without graphing
  • Solve maxima/minima problems efficiently
  • Understand curve sketching concepts deeply
  • Tackle differentiability questions confidently

1. f'(x) - The First Derivative: What It Tells Us

When f'(x) > 0

📈 Geometric Interpretation

The function is increasing at that point. The tangent line has positive slope.

🎯 JEE Example

For $f(x) = x^2$, $f'(x) = 2x$. When $x > 0$, $f'(x) > 0$, so the function is increasing.

Graph: Parabola opening upward, increasing for x > 0

When f'(x) < 0

📉 Geometric Interpretation

The function is decreasing at that point. The tangent line has negative slope.

🎯 JEE Example

For $f(x) = x^2$, $f'(x) = 2x$. When $x < 0$, $f'(x) < 0$, so the function is decreasing.

Graph: Parabola opening upward, decreasing for x < 0

When f'(x) = 0

⚖️ Geometric Interpretation

The function has a critical point (could be local max, local min, or point of inflection).

🎯 JEE Example

For $f(x) = x^2$, $f'(x) = 2x = 0$ when $x = 0$. This is a local minimum.

Graph: Parabola with minimum at (0,0)

2. f''(x) - The Second Derivative: Curvature Analysis

When f''(x) > 0

⎰ Geometric Interpretation

The function is concave up (convex). The curve lies above its tangent lines.

🎯 JEE Example

For $f(x) = x^2$, $f''(x) = 2 > 0$ for all x. The parabola is always concave up.

Graph: Concave up parabola

When f''(x) < 0

⎱ Geometric Interpretation

The function is concave down (concave). The curve lies below its tangent lines.

🎯 JEE Example

For $f(x) = -x^2$, $f''(x) = -2 < 0$ for all x. The parabola is always concave down.

Graph: Concave down parabola

When f''(x) = 0

🔄 Geometric Interpretation

Possible point of inflection - where concavity changes.

🎯 JEE Example

For $f(x) = x^3$, $f''(x) = 6x = 0$ when $x = 0$. This is a point of inflection.

Graph: Cubic curve with inflection at (0,0)

3. Geometric Relationships: Putting It All Together

The Complete Picture

f'(x) f''(x) Function Behavior JEE Significance
Positive Positive Increasing, Concave Up Rapid growth, exponential-like behavior
Positive Negative Increasing, Concave Down Slowing growth, approaching horizontal asymptote
Negative Positive Decreasing, Concave Up Slowing decrease, approaching minimum
Negative Negative Decreasing, Concave Down Rapid decrease, exponential decay

💡 Memory Aid

"Positive f'(x) means going up, Positive f''(x) means smiling up"

Think of f''(x) > 0 as a smile (concave up) and f''(x) < 0 as a frown (concave down).

4. Differentiability: Key Conditions and Common Pitfalls

When is a Function Differentiable?

✅ Differentiable at x = a if:

  • f(x) is continuous at x = a
  • Left-hand derivative = Right-hand derivative
  • No sharp corners or cusps
  • No vertical tangents

❌ NOT Differentiable at x = a if:

  • Discontinuity at x = a
  • Sharp corner (like |x| at x = 0)
  • Vertical tangent (like $x^{1/3}$ at x = 0)
  • Cusp point

⚠️ Common JEE Trap

Continuity ≠ Differentiability! A function can be continuous but not differentiable.

Example: $f(x) = |x|$ is continuous at x = 0 but not differentiable there (sharp corner).

🎯 JEE Differentiability Problem

Question: Check differentiability of $f(x) = \begin{cases} x^2 & \text{if } x \leq 1 \\ 2x-1 & \text{if } x > 1 \end{cases}$ at x = 1

Step 1: Check continuity at x = 1

Left limit: $\lim_{x \to 1^-} x^2 = 1$

Right limit: $\lim_{x \to 1^+} (2x-1) = 1$

f(1) = 1. Continuous ✓

Step 2: Check differentiability

Left derivative: $f'(x) = 2x \Rightarrow f'(1^-) = 2$

Right derivative: $f'(x) = 2 \Rightarrow f'(1^+) = 2$

Equal derivatives ✓

Conclusion: Differentiable at x = 1

5. JEE Applications and Problem Solving Strategies

Maxima/Minima Problems

Second Derivative Test

If f'(c) = 0 and
f''(c) > 0
Local Minimum at x = c
If f'(c) = 0 and
f''(c) < 0
Local Maximum at x = c
If f'(c) = 0 and
f''(c) = 0
Test inconclusive

Curve Sketching Strategy

1

Find critical points (f'(x) = 0)

2

Determine increasing/decreasing intervals

3

Find inflection points (f''(x) = 0)

4

Determine concavity intervals

🚀 Time-Saving JEE Strategy

For multiple choice questions: Often you can eliminate options by checking just the signs of f'(x) and f''(x) without full calculations.

Example: If asked about local maxima/minima, immediately check if f''(x) is positive or negative at critical points.

📋 Quick Reference Table

Function Property f'(x) Condition f''(x) Condition Geometric Meaning
Increasing f'(x) > 0 - Slope positive
Decreasing f'(x) < 0 - Slope negative
Local Maximum f'(x) = 0 f''(x) < 0 Peak point
Local Minimum f'(x) = 0 f''(x) > 0 Valley point
Concave Up - f''(x) > 0 Smile shape
Concave Down - f''(x) < 0 Frown shape

📝 Test Your Understanding

Try these JEE-style problems to apply what you've learned:

1. For $f(x) = x^3 - 3x^2 + 2$, find intervals where:

a) f(x) is increasing

b) f(x) is concave up

c) Local maxima and minima

2. Determine if $f(x) = |x-2| + |x+1|$ is differentiable at x = 2

3. A function has f'(1) = 0 and f''(1) = -3. What can you conclude about f(x) at x = 1?

Mastered the Relationships?

You're now ready to tackle any JEE question involving f(x), f'(x), and f''(x) relationships!

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