1. Given function

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

   To determine where the function is increasing or decreasing, we compute its derivative.

2. Differentiate

   $f'(x)=\frac{d}{dx}(x^3-3x^2+5x+7)=3x^2-6x+5$

3. Analyze the sign of $f'(x)$

   Rewrite:

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

   Check whether this quadratic can ever be negative.

   Its discriminant is:

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

   Since $\Delta<0$ and the coefficient of $x^2$ is positive ($3>0$), the quadratic $3x^2-6x+5$ is always positive for all real $x$.

   Hence,

   $f'(x)>0 \quad \forall x\in \mathbb{R}$

4. Conclusion about monotonicity

   Since $f'(x)>0$ for every real number, the function $f(x)$ is increasing on all of $\mathbb{R}$.

5. Option check

   • A: increasing in $\mathbb{R}$ — Correct
   • B: decreasing in $\mathbb{R}$ — Incorrect
   • C: decreasing in $(0,\infty)$ and increasing in $(-\infty,0)$ — Incorrect
   • D: increasing in $(0,\infty)$ and decreasing in $(-\infty,0)$ — Incorrect

Therefore, the correct answer is:

$\boxed{A}$

6. Comparison with stored answer

   Stored correct answer: A

   My derived answer also is A, so they agree.