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The function f(x)=x3+3x2+4x+4f(x) = x^3 + 3x^2 + 4x + 4, x∈Rx \in \mathbb{R} (set of real numbers) :

Solution

✅ Correct Option: 1

To determine if the function is increasing or decreasing, the derivative f′(x)f'(x) is examined. If f′(x)>0f'(x) > 0 for all xx, the function is increasing. If f′(x)<0f'(x) < 0 for all xx, the function is decreasing.


f(x)=x3+3x2+4x+4f(x) = x^3 + 3x^2 + 4x + 4

Using the power rule:

f′(x)=3x2+6x+4f'(x) = 3x^2 + 6x + 4


To determine the sign of f′(x)=3x2+6x+4f'(x) = 3x^2 + 6x + 4, the discriminant is calculated.

For the quadratic ax2+bx+cax^2 + bx + c where a=3a = 3, b=6b = 6, c=4c = 4:

D=b2−4acD = b^2 - 4ac

D=(6)2−4(3)(4)D = (6)^2 - 4(3)(4)

D=36−48D = 36 - 48

D=−12D = -12


Since D<0D < 0, the quadratic f′(x)=3x2+6x+4f'(x) = 3x^2 + 6x + 4 has no real roots.

Since the coefficient of x2x^2 is positive (a=3>0a = 3 > 0), the parabola opens upward.

Therefore, f′(x)>0f'(x) > 0 for all x∈Rx \in \mathbb{R}.


Since f′(x)>0f'(x) > 0 for all x∈Rx \in \mathbb{R}, the function f(x)=x3+3x2+4x+4f(x) = x^3 + 3x^2 + 4x + 4 is increasing on R\mathbb{R}.

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