Exercise 1 | Wiskunde op Tilburg University

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

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.