The function $$f$$ defined by $$f(x) = x^3 - 3x^2 + 5x + 7$$ is:

For any polynomial function, its monotonic behaviour on the real line can be studied by examining the sign of its first derivative. Let

$$f(x)=x^3-3x^2+5x+7 \; -(1)$$

First, differentiate term by term:

$$\frac{d}{dx}\bigl(x^3\bigr)=3x^2,\qquad \frac{d}{dx}\bigl(-3x^2\bigr)=-6x,\qquad \frac{d}{dx}\bigl(5x\bigr)=5,\qquad \frac{d}{dx}\bigl(7\bigr)=0.$$

Adding these results gives the derivative

$$f'(x)=3x^2-6x+5 \; -(2)$$

The derivative is a quadratic expression. Its sign on the entire real line can be found by checking its discriminant. For a general quadratic $$ax^2+bx+c,$$ the discriminant is $$D=b^2-4ac.$$ Here, $$a=3,\; b=-6,\; c=5.$$ Therefore

$$D=(-6)^2-4(3)(5)=36-60=-24 \; -(3)$$

The discriminant is negative, $$D\lt 0.$$ Because the leading coefficient $$a=3$$ is positive, the quadratic $$3x^2-6x+5$$ never changes sign and remains strictly positive for every real value of $$x$$. Symbolically,

$$3x^2-6x+5\gt 0\quad\text{for all }x\in\mathbb R \; -(4)$$

Equation $$-(4)$$ tells us that

$$f'(x)\gt 0\quad\forall x\in\mathbb R \; -(5)$$

Whenever the first derivative is positive everywhere, the original function is strictly increasing everywhere. Hence $$f(x)$$ increases for every real input, and it never decreases on any part of the real number line.

Hence, the correct answer is Option B.