Question bank — Increasing and decreasing functions — intuitive definition

False — the two pieces can jump. $f(x)=x$ for $x<0$ but $f(x)=x-5$ for $x\ge 0$ rises on each side yet drops across $0$, so a left point can sit above a right point.

True — strict increase forces distinct outputs for distinct inputs, since one of $x_1<x_2$ or $x_2<x_1$ holds and each gives a strict inequality on the outputs. This is why strictly monotone functions are one-to-one (see Inverse functions).

False — it is neither strictly increasing nor strictly decreasing, because equal outputs violate both the strict "$<$" and strict "$>$". It qualifies only as non-increasing and non-decreasing.

True — negating flips every output inequality: $f(x_1)<f(x_2)$ becomes $-f(x_1)>-f(x_2)$, which is exactly the decreasing condition.

True — a flat instant does not create a dip. For any $x_1<x_2$ we still get $x_1^3<x_2^3$, so order is preserved everywhere; being "increasing" is about order, not about the slope being positive at every single point.

False — products need not preserve order. On $(-\infty ,0)$ take $f(x)=g(x)=x$ (both increasing); their product $x^2$ is decreasing there.

True — adding two order-preserving quantities keeps order: $x_1<x_2$ gives $f(x_1)<f(x_2)$ and $g(x_1)<g(x_2)$, and summing the inequalities gives $(f+g)(x_1)<(f+g)(x_2)$.

True — "increasing" describes the direction of change, not the sign of the output. $f(x)=x-10$ is strictly increasing yet negative for all $x<10$.

True — since a strictly decreasing $f$ is one-to-one, its inverse exists, and reflecting a downhill graph across $y=x$ leaves it downhill; the inverse is decreasing too.

The error is testing one pair. The definition requires the inequality for all pairs in the interval; a single successful pair can coexist with failures elsewhere, as $x^2$ shows with $f(-2)>f(-1)$.

There is no "more increasing." Both preserve order everywhere, so both are simply increasing; the difference is steepness, which is a rate-of-change question, not a monotonicity one.

The claim ignores the gap at $0$. Take $x_1=-1<x_2=1$: then $f(-1)=-1<1=f(1)$, an increase across the gap. It is decreasing on $(-\infty ,0)$ and on $(0,\infty )$ separately, never as one interval — domain matters.

The "nowhere monotone" part is wrong. Equal outputs at $-2$ and $2$ only block monotonicity on an interval containing both; on $(0,\infty )$ it is strictly increasing and perfectly one-to-one.

A proof needs the difference to be positive for general $x_1<x_2$, not chosen numbers. Keep $x_1,x_2$ as symbols and show $f(x_2)-f(x_1)>0$ follows from $x_2-x_1>0$.

Direction convention is broken. "Increasing" always means as $x$ grows (reading left to right) the height grows; reading right-to-left reverses the sense and would call a decreasing function increasing.

Passing the horizontal-line test only shows $f$ is one-to-one, which permits either pure rise or pure fall (or worse, a mix with a jump). One-to-one is necessary for strict monotonicity but not sufficient by itself.

Because monotonicity is a global-order property on the interval; a function could rise between close points yet still let a far-left point exceed a far-right one if we didn't demand every pair, so "all pairs" seals every loophole at once.

Distinct inputs are ordered one way or the other, and each ordering forces a strict output inequality, so two different inputs can never share an output — exactly the one-to-one condition.

Steepness measures how fast the output changes, but increasing only asks in which direction it changes; a barely-rising curve and a steep one are equally "increasing," so steepness is the wrong measuring stick.

Most functions change behaviour across their domain (rise here, fall there), so a bare "increasing" is ambiguous; naming the interval pins down exactly where the order-preserving claim is being made.

The sign of that single difference answers the whole question — positive means the right point is higher (increasing), negative means lower (decreasing) — converting a comparison into a clean sign analysis.

Non-decreasing allows flat stretches where $f(x_1)=f(x_2)$, so a staircase or constant piece qualifies; strictly increasing forbids equality, demanding a real climb between every distinct pair.

Vacuously — there are no two distinct points to compare, so "for all pairs $x_1<x_2$" is trivially satisfied; such a function is increasing, decreasing, and constant all at once by emptiness of the condition.

Yes — an isolated zero-slope point (like $x^3$ at $0$) causes no reversal, so strict order across the interval survives; monotonicity is judged by outputs, not by the derivative at one instant.

Yes — the definition only quantifies over points in the interval, so a missing endpoint is irrelevant; $\frac{1}{x}$-style domains work fine on open intervals as long as no gap sits inside them.

No — it oscillates, rising on some intervals and falling on others, so it fails the all-pairs test globally; monotonicity for it must always be stated on a chosen sub-interval like $[-\frac{\pi }{2},\frac{\pi }{2}]$.

Yes — $a$ and $b$ are themselves a valid pair with $a<b$, so the definition directly forces $f(a)<f(b)$; the minimum sits at the left end and the maximum at the right.