Answer
$f$ is increasing for $x=1$.
Work Step by Step
Recall that:
The function$f$ is increasing if:
$x \gt y \to f(x) \gt f(y)$
The function$f$ is decreasing if:
$x \gt y \to f(x) \lt f(y)$
The instantaneous rate of change of $f$ at $x=1$ is:
$$\lim\limits_{h \to 0}\frac{f(1+h)-f(1)}{h} \gt 0$$
It follows that the rate of change is also positive:
$$\frac{f(1+h)-f(1)}{h} \gt 0$$
Since the step size $h$ is always positive it follows $h\gt 0$ or $1+h \gt 1$ so:
$$f(1+h)-f(1) \gt 0$$
$$f(1+h)\gt f(1) $$
Since $1+h \gt 1$ we have $f(1+h) \gt f(1)$ it follows that $f$ is increasing for $x=1$.