Find Local Maximum and Minimum of f(x) = x³ - 3x + 1

medium CBSE JEE-MAIN CBSE 2024 Board Exam 4 min read

Question

Find all local maxima and minima of f(x) = x³ - 3x + 1. Also state the local maximum and minimum values.

This question appeared in the CBSE 2024 Board Exam and is a standard first derivative test problem. Full marks require both the x-coordinates AND the function values.


Solution — Step by Step

Differentiate f(x) = x³ - 3x + 1 term by term:

f(x)=3x23f'(x) = 3x^2 - 3

This tells us the slope of the curve at every point. Where slope = 0, we have a candidate for extremum.

3x23=0    x2=1    x=±13x^2 - 3 = 0 \implies x^2 = 1 \implies x = \pm 1

We get two critical points: x = 1 and x = -1. Both must be tested — don’t discard either.

f(x)=6xf''(x) = 6x

The second derivative test tells us: if f''(x) > 0 at a critical point, it’s a local minimum; if f''(x) < 0, it’s a local maximum.

At x = 1:

f(1)=6(1)=6>0    Local Minimumf''(1) = 6(1) = 6 > 0 \implies \text{Local Minimum}

At x = -1:

f(1)=6(1)=6<0    Local Maximumf''(-1) = 6(-1) = -6 < 0 \implies \text{Local Maximum}

Local minimum value (at x = 1):

f(1)=(1)33(1)+1=13+1=1f(1) = (1)^3 - 3(1) + 1 = 1 - 3 + 1 = -1

Local maximum value (at x = -1):

f(1)=(1)33(1)+1=1+3+1=3f(-1) = (-1)^3 - 3(-1) + 1 = -1 + 3 + 1 = 3

Final Answer: Local maximum = 3 at x = -1; Local minimum = -1 at x = 1.


Why This Works

The function f(x) = x³ - 3x + 1 is a cubic polynomial, and all cubics have an S-shape. The curve rises steeply, flattens out at a “hump” (local max), dips down to a “valley” (local min), then rises steeply again. The derivative f'(x) captures where this flattening happens — it’s zero exactly at these turning points.

The second derivative test works because f''(x) measures the rate of change of slope. At x = -1, the slope is going from positive (rising) to zero to negative (falling) — that’s a peak. Since the slope is decreasing there, f'' is negative. At x = 1, slope goes from negative to zero to positive — that’s a valley, so f'' is positive.

This logic holds universally: positive f'' means the curve is “concave up” like a bowl (holds water), which is always a minimum. Negative f'' means “concave down” like an arch, which is always a maximum.


Alternative Method

We can use the first derivative sign change test instead — useful when f''(c) = 0 (second derivative test fails).

Check the sign of f'(x) = 3(x² - 1) = 3(x-1)(x+1) in three intervals:

IntervalSign of (x-1)Sign of (x+1)Sign of f'(x)
x < -1+ (increasing)
-1 < x < 1+− (decreasing)
x > 1+++ (increasing)

At x = -1: sign changes from + to Local Maximum At x = 1: sign changes from to +Local Minimum

Same conclusion, no second derivative needed.

In CBSE board exams, both methods fetch full marks. But the second derivative test is faster when f”(c) ≠ 0 — use it by default and switch to sign change only when f”(c) = 0.


Common Mistake

Stopping after finding critical points. Many students write “x = 1 is minimum and x = -1 is maximum” and stop there — losing 1-2 marks. The question asks for local maximum and minimum values. You must substitute back: f(-1) = 3 and f(1) = -1. Always re-read whether the question wants the x-coordinate (point) or the y-value (value of function).

A second common slip: computing f(-1) as (-1)³ = 1 instead of -1. Odd powers of negative numbers stay negative. (-1)³ = -1, not +1.

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