Application of Derivatives: Conceptual Doubts Cleared (2)

medium 3 min read

Question

Find the local maxima, local minima, and points of inflection of f(x)=x36x2+9x+1f(x) = x^3 - 6x^2 + 9x + 1 on the real line. State the intervals of increase and decrease. CBSE 2024 boards (6-mark) and JEE Main pattern.

Solution — Step by Step

f(x)=3x212x+9=3(x24x+3)=3(x1)(x3)f'(x) = 3x^2 - 12x + 9 = 3(x^2 - 4x + 3) = 3(x - 1)(x - 3)

Critical points: x=1x = 1 and x=3x = 3.

f(x)=6x12f''(x) = 6x - 12

At x=1x = 1: f(1)=6<0f''(1) = -6 < 0, so x=1x = 1 is a local maximum.

At x=3x = 3: f(3)=6>0f''(3) = 6 > 0, so x=3x = 3 is a local minimum.

Set f(x)=0f''(x) = 0: 6x12=0    x=26x - 12 = 0 \implies x = 2. Since ff'' changes sign at x=2x = 2 (negative for x<2x < 2, positive for x>2x > 2), this is a point of inflection.

f(2)=824+18+1=3f(2) = 8 - 24 + 18 + 1 = 3. So inflection point is (2,3)(2, 3).

f(x)>0f'(x) > 0 when (x1)(x3)>0(x - 1)(x - 3) > 0, i.e., x<1x < 1 or x>3x > 3. Increasing on (,1)(3,)(-\infty, 1) \cup (3, \infty).

f(x)<0f'(x) < 0 when 1<x<31 < x < 3. Decreasing on (1,3)(1, 3).

Function values: f(1)=16+9+1=5f(1) = 1 - 6 + 9 + 1 = 5 (local max value), f(3)=2754+27+1=1f(3) = 27 - 54 + 27 + 1 = 1 (local min value).

Why This Works

The first derivative tells us where the slope is zero (critical points) and the sign of the slope (increasing or decreasing). The second derivative tells us concavity — concave down means a maximum, concave up means a minimum.

Inflection points are where concavity changes. They’re not extrema; the function can be increasing or decreasing through them. JEE Main and CBSE both ask for inflection points separately from extrema, so don’t conflate them.

Alternative Method

The first derivative test (sign-change analysis) works without computing ff'':

f(x)=3(x1)(x3)f'(x) = 3(x-1)(x-3). For x<1x < 1: both factors negative, so f>0f' > 0. Between 1 and 3: (x1)>0(x-1) > 0, (x3)<0(x-3) < 0, so f<0f' < 0. For x>3x > 3: both positive, f>0f' > 0.

Sign changes: positive → negative at x=1x = 1 (local max), negative → positive at x=3x = 3 (local min). Same conclusion.

When f(c)=0f''(c) = 0, the second derivative test is inconclusive — you cannot conclude max, min, or saddle without more analysis. JEE Advanced loves to set up problems where f(c)=0f''(c) = 0 at a critical point precisely to trap students who only know the second-derivative test.

For cubics ax3+bx2+cx+dax^3 + bx^2 + cx + d with positive leading coefficient: if there are two critical points, the smaller is a local max, the larger is a local min. Saves time in MCQs.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next