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The interval in which the function f(x)=2x3+3x2−12x+1f(x) = 2x^3 + 3x^2 - 12x + 1 is strictly increasing, is

Solution

✅ Correct Option: 1

Given: f(x)=2x3+3x2−12x+1f(x) = 2x^3 + 3x^2 - 12x + 1

A function is strictly increasing when its derivative is positive.

f′(x)=6x2+6x−12f'(x) = 6x^2 + 6x - 12


For strictly increasing:

f′(x)>0f'(x) > 0

6x2+6x−12>06x^2 + 6x - 12 > 0

x2+x−2>0x^2 + x - 2 > 0


Factoring the quadratic:

x2+x−2=(x+2)(x−1)x^2 + x - 2 = (x + 2)(x - 1)

(x+2)(x−1)>0(x + 2)(x - 1) > 0


Critical points where each factor equals zero:

x+2=0⇒x=−2x + 2 = 0 \Rightarrow x = -2

x−1=0⇒x=1x - 1 = 0 \Rightarrow x = 1


Testing the sign of (x+2)(x−1)(x + 2)(x - 1) in each region:

For x<−2x < -2 (test x=−3x = -3): (−)(−)=(+)(-)(-) = (+) ✓

For −2<x<1-2 < x < 1 (test x=0x = 0): (+)(−)=(−)(+)(-) = (-) ✗

For x>1x > 1 (test x=2x = 2): (+)(+)=(+)(+)(+) = (+) ✓


The function is strictly increasing on (−∞,−2)∪(1,∞)(-\infty, -2) \cup (1, \infty)

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