Find the local maximum and local minimum values of $f(x) = x^3 - 3x$ using the first derivative test.

$f(x)=x^3-3x \implies f'(x)=3x^2-3=3(x-1)(x+1)$. Critical points: $x=-1,\ 1$. | Interval | Test point | $f'(x)$ sign | |---|---|---| | $x 0$ | | $-1 1$ | $x=2$ | $3(1)(3)=9>0$ | At $x=-1$: sign changes $+\to-$, so $f$ has a **local maximum**; $f(-1) = -1+3 = 2$. …

Find the local maximum and local minimum values of – Class 12 Mathematics

Example · Example 7

f(x) = x^3 - 3x

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$f(x)=x^3-3x \implies f'(x)=3x^2-3=3(x-1)(x+1)$ . Critical points: $x=-1,\ 1$ .

Interval	Test point	$f'(x)$ sign
$x<-1$	$x=-2$	$3(-3)(-1)=9>0$
$-1<x<1$	$x=0$	$3(-1)(1)=-3<0$
$x>1$	$x=2$	$3(1)(3)=9>0$

At $x=-1$ : sign changes $+\to-$ , so $f$ has a local maximum; $f(-1) = -1+3 = 2$ . …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set SEM31 markMCQ
$y = \dfrac{\log_e x}{x}$
(a) $e$
(b) $e^2$
(c) $\dfrac{1}{e}$
(d) $\dfrac{1}{e^2}$
›Reveal solutionSolution

Differentiate $y=\dfrac{\log_e x}{x}$ , set $y'=0$ to get $x=e$ , and evaluate; the maximum value is $\dfrac1e$ .

Maximising a function using the first derivative is a standard NCERT/CBSE Class 12 application of derivatives problem.

Differentiate with the quotient rule:

$y' = \frac{\tfrac{1}{x}\cdot x - \log_e x \cdot 1}{x^2} = \frac{1-\log_e x}{x^2}.$

…

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CBSE 2026Set SEM31 markMCQ
$f(x) = 3 + |x - 3|$
(a) Statement-I is true, Statement-II is false
(b) Statement-I is false, Statement-II is true
(c) Statements-I and II both are true
(d) Statements-I and II both are false
›Reveal solutionSolution

Both statements are correct, so the answer is (c).

These are qualitative NCERT/CBSE Class 12 maxima–minima facts.

Statement-I: $|x-3| \ge 0$ for all $x$ , with equality only at $x=3$ . Hence $f(x)=3+|x-3| \ge 3$ , and the least value $3$ is achieved at $x=3$ . So it has a local (indeed global) minimum value $3$ . True.

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. Maximum value of 5 - (x-1)² is$
(a) 5
(b) 4
(c) 6
(d) 3
›Reveal solutionSolution

$(x-1)^2 \ge 0$ always, so $5-(x-1)^2$ is largest exactly when the squared term is $0$ .

For any real $x$ , $(x-1)^2 \ge 0$ , with equality exactly when $x=1$ . So $5-(x-1)^2 \le 5$ for all $x$ , and this maximum value $5$ is attained at $x=1$ .

…

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CBSE 2022Set ANNUAL1 markMCQ
$Q. The value of the function f(x) = 4x - x² - 3 will be maximum when:$
(a) x = 3
(b) x = 2
(c) x = -2
(d) x = -3
›Reveal solutionSolution

Set the derivative to zero to find the critical point, then confirm it's a maximum with the second derivative.

Given $f(x)=4x-x^2-3$ .

First derivative:

$f'(x) = 4-2x$

Set $f'(x)=0$ :

$4-2x=0 \Rightarrow x=2$

Second derivative test: …

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Q.Find the local maximum and local minimum values of $f(x) = x^3 - 3x$ using the first derivative test.

Concept understanding — Maxima and Minima -- First Derivative Test

Q.Find the local maximum and local minimum values of f(x)=x3−3xf(x) = x^3 - 3xf(x)=x3−3x using the first derivative test.

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Q.Find the local maximum and local minimum values of $f(x) = x^3 - 3x$ using the first derivative test.

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