Collection of Maths Problems

Assertions about monotonicity and extremes

Task number: 2974

• Variant 1

  A function \(f\) is increasing on the interval \(I\) if and only if: \( \forall x,y\in I, x\ne y: \frac{f(x)-f(y)}{x-y}> 0 \)

• Resolution

  To say that \(f\) is increasing means that \(x>y \longrightarrow f(x)> f(y)\).

  Forward implication: \(x>y \longrightarrow (x-y>0) \land (f(x)-f(y)>0) \longrightarrow \frac{f(x)-f(y)}{x-y}> 0\).
  (Analogously for \(x<y\).)

  Reverse implication: \(\left(\frac{f(x)-f(y)}{x-y}> 0 \right)\land (x-y>0) \longrightarrow f(x)-f(y)>0\).

• Result

  The assertion holds.

• Variant 2

  If a function \(f\) is non-decreasing on the interval \((-\infty,a\rangle\) and non-increasing on \(\langle a,\infty)\) for some \(a\in\mathbb R\), then \(f\) has a maximum.

• Resolution

  Non-decreasing on \((-\infty,a\rangle \Leftrightarrow \forall b\in (-\infty,a): f(b)\le f(a)\).

  Non-increasing on \(\langle a,\infty) \Leftrightarrow \forall b\in (a,\infty): f(b)\le f(a)\).

  So \(\forall b\in \mathbb R: f(b)\le f(a)\), and also \(f(a)\in \text{H}(f)\), so \(f(a)=\max \text{H}(f)\).

• Result

  The assertion holds.

• Variant 3

  If the function \(f\) has a minimum at the point \(a\in\mathbb R\), then there exists \(\varepsilon>0\) such that \(f\) is non-increasing on the interval \((a-\varepsilon,a]\) and non-decreasing on \([ a,a+\varepsilon)\).

• Resolution

  A counterexample is the function defined as \(f(x)=\cos x\) for \(x\ne 0\) and \(f(0)=-2\), which has a minimum.

  For any \(\varepsilon>0\), \(f\) is increasing on \((-\varepsilon,0)\), and decreasing on \((0,\varepsilon)\).

  Note: the assertion would not hold even assuming continuity, for example for the function \(f(x)=|x|+\frac{|x|}3\sin\frac1x\).

• Result

  The assertion does not hold.

• Variant 4

  If the function \(f\) has limit \(\infty\) at point \(c\), then the value of the function \(f\) on every punctured neighborhood \(P_\delta(c)\) is unlimited.

• Resolution

  \( \displaystyle \lim_{x\to c}f(x)=\infty \Leftrightarrow (\forall a \exists \delta_a: |c-x|< \delta_a \longrightarrow f(x)>a) \)

  So for any punctured neighborhood \(P_\rho(c)\) and an arbitrarily large \(a\), then for \(x=c+\frac12 \min\{\rho,\delta_a\}\) we have \(x\in P_\rho(c)\) and \(f(x)>a\).

• Result

  The assertion holds.

• Variant 5

  If for every punctured neighborhood \(P_\delta(c)\) the value of the function \(f\) is not bounded above, then \(\displaystyle \lim_{x\to c}f(x)=\infty \).

• Resolution

  This is not true, because the limit might not exist, e.g. for the function \(f(x)=\frac1x\sin{\frac1x}\) in the neighborhood of 0.

• Result

  The assertion does not hold.