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  2. For business economics
  3. Chapter 5: Optimization
  4. Optimization functions of one variable
  5. Alternative monotonicity condition extremum
  6. Exercise 1

Exercise 1

Determine all the extrema of $y(x)=4x^2-16x+7$.
$y(2)=-9$ is a minimum.
$y(2)=-9$ is a maximum.
$y(\frac{1}{2})=0$ is a maximum.
$y(\frac{1}{2})=0$ is a minimum.
Determine all the extrema of $y(x)=4x^2-16x+7$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$y(2)=-9$ is a maximum.
Antwoord 2 correct
Fout
Antwoord 3 optie
$y(\frac{1}{2})=0$ is a maximum.
Antwoord 3 correct
Fout
Antwoord 4 optie
$y(\frac{1}{2})=0$ is a minimum.
Antwoord 4 correct
Fout
Antwoord 1 optie
$y(2)=-9$ is a minimum.
Antwoord 1 feedback
Correct: $y'(x)=8x-16$. Hence, $y'(x)=0$ if $x=2$. Since $y(2)=-9$, $y(0)=7$, $y(4)=7$ it holds that $y(2)=-9$ is a minimum.

Go on.
Antwoord 2 feedback
Wrong: Is this a maximum?

See Alternative monotonicity condition extremum.
Antwoord 3 feedback
Wrong: What is the solution to $8x=16$?

Try again.
Antwoord 4 feedback
Wrong: What is the solution to $8x=16$?

Try again.