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  3. Chapter 5: Optimization
  4. Optimization functions of one variable
  5. Alternative monotonicity condition extremum
  6. Exercise 3

Exercise 3

Determine all the extrema of $y(x)=\textrm{ln}(x)-2x$.
$y(\frac{1}{2})=\textrm{ln}(\frac{1}{2})-1$ is a maximum.
  • $y(\frac{1}{2})=\textrm{ln}(\frac{1}{2})-1$ is a maximum.
  • $y(0)=-\infty$ is a boundary minimum.
  • $y(\frac{1}{2})=\textrm{ln}(\frac{1}{2})-1$ is a minimum.
  • $y(0)=0$ is a boundary maximum.
$y(\frac{1}{2})=\textrm{ln}(\frac{1}{2})-1$ is a minimum.
Determine all the extrema of $y(x)=\textrm{ln}(x)-2x$.
Antwoord 1 correct
Correct
Antwoord 2 optie
  • $y(\frac{1}{2})=\textrm{ln}(\frac{1}{2})-1$ is a maximum.
  • $y(0)=-\infty$ is a boundary minimum.
Antwoord 2 correct
Fout
Antwoord 3 optie
  • $y(\frac{1}{2})=\textrm{ln}(\frac{1}{2})-1$ is a minimum.
  • $y(0)=0$ is a boundary maximum.
Antwoord 3 correct
Fout
Antwoord 4 optie
$y(\frac{1}{2})=\textrm{ln}(\frac{1}{2})-1$ is a minimum.
Antwoord 4 correct
Fout
Antwoord 1 optie
$y(\frac{1}{2})=\textrm{ln}(\frac{1}{2})-1$ is a maximum.
Antwoord 1 feedback
Correct: $y'(x)=\frac{1}{x}-2$ and hence, $y'(x)=0$ for $x=\frac{1}{2}$. Since $y(\frac{1}{2})=\textrm{ln}(\frac{1}{2})-1$ , $y(\frac{1}{4})=\textrm{ln}(\frac{1}{4})-\frac{1}{2}$ and $y(1)=-2$ it holds that $y(\frac{1}{2})=\textrm{ln}(\frac{1}{2})-1$ is a maximum.

Go on.
Antwoord 2 feedback
Wrong: $x=0$ is outside the domain of the function ln$(x)$. Hence, there is no boundary extremum.

See Extra explanation: natural logarithm.
Antwoord 3 feedback
Wrong: $x=0$ is outside the domain of the function ln$(x)$. Hence, there is no boundary extremum.

See Extra explanation: natural logarithm.
Antwoord 4 feedback
Wrong: Calculate $y(\frac{1}{2})$, $y(\frac{1}{4})$ and $y(1)$.

See Alternative monotonicity condition extremum.