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

Exercise 2

Determine all stationary points of $f(x)=x^3-4x^2+6x$.
There are no stationary points.
$x=4+\frac{1}{2}\sqrt{40}$ and $x=4-\frac{1}{2}\sqrt{40}$.
$x=0$ and $x=2\frac{2}{3}$
$x=0$
Determine all stationary points of $f(x)=x^3-4x^2+6x$.
Antwoord 1 correct
Correct
Antwoord 2 optie
$x=4+\frac{1}{2}\sqrt{40}$ and $x=4-\frac{1}{2}\sqrt{40}$.
Antwoord 2 correct
Fout
Antwoord 3 optie
$x=0$ and $x=2\frac{2}{3}$
Antwoord 3 correct
Fout
Antwoord 4 optie
$x=0$
Antwoord 4 correct
Fout
Antwoord 1 optie
There are no stationary points.
Antwoord 1 feedback
Correct: $f'(x)=3x^2-8x+6$. $D=(-8)^2-4\cdot 3 \cdot 6=-8<0$ and hence, $f'(x)$ has no zeros, which implies that $f(x)$ has no stationary points.

Go on.
Antwoord 2 feedback
Wrong: $f'(x) \neq x^2-8x+6$.

See Derivative.
Antwoord 3 feedback
Wrong: The discriminant $D$ is equal to $D=b^2-4ac$, not to $D=b^2+4ac$.

See Extra explanation: zeros.
Antwoord 4 feedback
Wrong: A stationairy point $c$ is not a point such that $y(c)=0$.

See Stationary point.