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  2. For business economics
  3. Chapter 5: Optimization
  4. Optimization functions of one variable
  5. Stationary point
  6. Exercise 1

Exercise 1

Determine all stationary points of $y(x)=-x^2+6x+7$.
$x=3$
$x=-1$ and $x=7$
$x=6\frac{1}{2}$
$y(x)$ has no stationary points.
Determine all stationary points of $y(x)=-x^2+6x+7$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$x=-1$ and $x=7$
Antwoord 2 correct
Fout
Antwoord 3 optie
$x=6\frac{1}{2}$
Antwoord 3 correct
Fout
Antwoord 4 optie
$y(x)$ has no stationary points.
Antwoord 4 correct
Fout
Antwoord 1 optie
$x=3$
Antwoord 1 feedback
Correct: $y'(x)=-2x+6$. Hence, $y'(x)=0$ gives $x=3$. Therefore, $x=3$ is the unique stationary point of $y(x)$.

Go on.
Antwoord 2 feedback
Wrong: A stationary point $c$ is not a point such that $y(c)=0$.

See Stationary point.
Antwoord 3 feedback
Wrong: $y'(x)\neq -2x+6+7$.

See Derivative.
Antwoord 4 feedback
Wrong: $y(x)$ does have a stationary point.

See Example.