Exercise 2 | Wiskunde op Tilburg University

Determine all stationary points of

$f(x)=x^3-4x^2+6x$ .

$x=4+\frac{1}{2}\sqrt{40}$ and

$x=4-\frac{1}{2}\sqrt{40}$ .

$x=0$

$x=0$ and

$x=2\frac{2}{3}$

There are no stationary points.

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.