15. Given the function
$$
f(x)=x^{3}-m x^{2}-x+1(m \in \mathbf{R}) \text {. }
$$
(1) Find the monotonic intervals of the function $f(x)$;
(2) If for all real numbers $x$, we have
$$
f^{\prime}(x) \geqslant|x|-\frac{7}{4}
$$

holds, find the range of the real number $m$.
15. (1) Since $f^{\prime}(x)=3 x^{2}-2 m x-1$, and
$$
\Delta=4 m^{2}+12>0 \text {, }
$$

therefore, $f^{\prime}(x)=0$ has two distinct real roots
$$
x_{1}=\frac{m-\sqrt{m^{2}+3}}{3}, x_{2}=\frac{m+\sqrt{m^{2}+3}}{3} .
$$

Clearly, $x_{1}x_{2}$ or $x0$, i.e., $f(x)$ is monotonically increasing.
In summary, the monotonically decreasing interval of $f(x)$ is
$$
\left[\frac{m-\sqrt{m^{2}+3}}{3}, \frac{m+\sqrt{m^{2}+3}}{3}\right] ;
$$

the monotonically increasing intervals are
$$
\left(-\infty, \frac{m-\sqrt{m^{2}+3}}{3}\right),\left(\frac{m+\sqrt{m^{2}+3}}{3},+\infty\right) \text {. }
$$
(2) According to the problem, we have
$$
3 x^{2}-2 m x-1 \geqslant|x|-\frac{7}{4} \text {. }
$$
(i) When $x>0$, we have
$$
3 x^{2}-(2 m+1) x+\frac{3}{4} \geqslant 0 \text {, }
$$

which means $3 x+\frac{3}{4 x} \geqslant 2 m+1$
holds for all $x>0$.
Notice that
$$
3 x+\frac{3}{4 x} \geqslant 2 \sqrt{3 x \cdot \frac{3}{4 x}}=3 \text {. }
$$

The equality holds when $x=\frac{1}{2}$.
Therefore, $3 \geqslant 2 m+1 \Rightarrow m \leqslant 1$.
(ii) When $x<0$, we have
$$
3|x|^{2}+(2 m-1)|x|+\frac{3}{4} \geqslant 0 \text {, }
$$

which means $3|x|+\frac{3}{4|x|} \geqslant 1-2 m$
holds for all $x<0$.
Notice that
$$
3|x|+\frac{3}{4|x|} \geqslant 2 \sqrt{3|x| \cdot \frac{3}{4|x|}}=3 .
$$

The equality holds when $x=-\frac{1}{2}$.
Therefore, $3 \geqslant 1-2 m \Rightarrow m \geqslant-1$.
(iii) When $x=0$, $m \in \mathbf{R}$.

In summary, the range of real number $m$ is $[-1,1]$.