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JEE Mains Maths · Application of Derivatives

Greatest and Least Values

The absolute maximum and minimum of a function: on a closed interval by comparing critical points with the endpoints, and over a whole domain by limits, substitution or AM-GM.

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Why this matters

Twenty-six PYQs, twenty-one of them multiple choice, and four from 2026. Eighteen find the greatest and least values on a closed interval, often with a modulus or a greatest-integer part that adds corners; eight find a range or a least value over an open domain, where a substitution or AM-GM is quicker than differentiating. Two ideas cover the page.

Concept 1 of 2: Closed interval: critical points and endpoints

On [a,b][a,b], a continuous function reaches its greatest and least values either at an endpoint or at a critical point inside. So list the critical points in the interval, add aa and bb, evaluate ff at all of them, and pick the largest and smallest. If ff is monotonic on the interval, only the endpoints matter. For ∣g∣|g| or [g][g], split the interval where gg changes sign or crosses an integer.

Definition

  • Candidates: endpoints, zeros of f′f', points where f′f' does not exist.
  • Monotonic on [a,b][a,b]: extremes at aa and bb.
  • tan⁡−1\tan^{-1}, exe^x, ln⁡x\ln x are increasing: extremes of tan⁡−1g\tan^{-1}g come from extremes of gg.
  • [g][g] is constant between the points where gg crosses an integer.

Closed-interval method

max⁡[a,b]f=max⁡{f(a), f(b), f(c):f′(c)=0}\max_{[a,b]}f=\max\{f(a),\,f(b),\,f(c):f'(c)=0\}

Worked example

Find the greatest and least values of x3−3xx^3-3x on [0,2][0,2].
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 30 Jan 2024 · Q155Moderate

Example 1 · Application of Derivatives · Greatest and Least Values

Let f(x)=(x+3)2(x−2)3,x∈[−4,4]f(x) = (x+ 3)^{2}(x- 2)^{3},x\in \lbrack - 4,4\rbrack. If MM and mm are the maximum and minimum values of ff, respectively in [−4,4]\lbrack - 4,4\rbrack, then the value of M−mM-m is:

The endpoints count

A local maximum inside the interval can be smaller than the value at an endpoint. In (x+3)2(x−2)3(x+3)^2(x-2)^3 on [−4,4][-4,4], both the greatest and the least values sit at endpoints.
Drill 17 more on closed interval: critical points and endpoints

Concept 2 of 2: Ranges and least values over a whole domain

Over an open domain there are no endpoints, so compare the critical values with the limits at the ends of the domain. Two shortcuts save work. For a ratio of quadratics, set it equal to yy and require the quadratic in xx to have a real root. For a sum of two positive terms whose product is fixed, AM-GM gives the least value directly.

Definition

  • Compare critical values with the limits at the ends of the domain.
  • y=P(x)Q(x)y=\frac{P(x)}{Q(x)} (quadratics): real xx needs a discriminant ≥0\ge0 in xx.
  • AM-GM: a+b≥2aba+b\ge2\sqrt{ab} for a,b>0a,b>0, equality at a=ba=b.
  • Substitute t=sin⁡xt=\sin x, t=ext=e^x and so on, keeping tt in its range.

AM-GM

a+b≥2ab(a,b>0), equality when a=ba+b\ge2\sqrt{ab}\quad(a,b>0),\ \text{equality when }a=b

Worked example

Find the least value of x2+16x2x^2+\frac{16}{x^2}.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q65Moderate

Example 2 · Application of Derivatives · Greatest and Least Values

Let the sum of the maximum and the minimum values of the function f(x)=2x2−3x+82x2+3x+8f(x) =\frac{2x^{2}- 3x + 8}{2x^{2}+ 3x + 8} be mn\frac{m}{n}, where gcd(m,n)=1gcd(m,n) = 1. Then m+nm + n is equal to:

Is the bound reached?

AM-GM and the discriminant give bounds. Check that the equality case happens at an allowed xx; if it does not, the bound is not the least value.
Drill 7 more on ranges and least values over a whole domain

Summary — formulas & gotchas at a glance

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Test yourself on Application of Derivatives

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