CUET UG 2025 — Mathematics Calculus

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

A function is decreasing when its derivative is negative.

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Finding the derivative using the power rule:

$$f'(x) = 3x^2 + 4x$$f′(x)=3x2+4x

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To find critical points, set $f'(x) = 0$f′(x)=0:

$$3x^2 + 4x = 0$$3x2+4x=0

$$x(3x + 4) = 0$$x(3x+4)=0

This gives $x = 0$x=0 or $3x + 4 = 0$3x+4=0

Therefore $x = 0$x=0 or $x = -\dfrac{4}{3}$x=−34​

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Testing the sign of $f'(x) = x(3x + 4)$f′(x)=x(3x+4) in each interval:

For $x < -\dfrac{4}{3}$x<−34​, test $x = -2$x=−2:

$$f'(-2) = (-2)(3(-2) + 4) = (-2)(-2) = 4 > 0$$f′(−2)=(−2)(3(−2)+4)=(−2)(−2)=4>0

For $-\dfrac{4}{3} < x < 0$−34​<x<0, test $x = -1$x=−1:

$$f'(-1) = (-1)(3(-1) + 4) = (-1)(1) = -1 < 0$$f′(−1)=(−1)(3(−1)+4)=(−1)(1)=−1<0

For $x > 0$x>0, test $x = 1$x=1:

$$f'(1) = (1)(3(1) + 4) = (1)(7) = 7 > 0$$f′(1)=(1)(3(1)+4)=(1)(7)=7>0

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The function is decreasing where $f'(x) < 0$f′(x)<0, which occurs in the interval $-\dfrac{4}{3} < x < 0$−34​<x<0.

In interval notation: $\left[-\dfrac{4}{3}, 0\right)$[−34​,0)

The bracket at $-\dfrac{4}{3}$−34​ includes the endpoint, while the parenthesis at $0$0 excludes it.

Therefore, the function is decreasing on $\left[-\dfrac{4}{3}, 0\right)$[−34​,0).