Exercise 3 | Wiskunde op Tilburg University

Determine all the extrema of

$y(x)=(x-2)e^{x^2}$ .

Antwoord 1 correct

Correct

Antwoord 2 optie

$y(0)=-2$ is a minimum

Antwoord 2 correct

Fout

Antwoord 3 optie

Antwoord 3 correct

Fout

Antwoord 4 optie

Antwoord 4 correct

Fout

Antwoord 1 optie

$y(1-\frac{1}{2}\sqrt{2})=(-1-\frac{1}{2}\sqrt{2})e^{1\frac{1}{2}-\sqrt{2}}$ is a maximum
$y(1+\frac{1}{2}\sqrt{2})=(-1+\frac{1}{2}\sqrt{2})e^{1\frac{1}{2}+\sqrt{2}}$ is a minimum

Antwoord 1 feedback

Correct:

$y'(x)=e^{x^2}+2x(x-2)e^{x^2}=(2x^2-4x+1)e^{x^2}$ .

$y'(x)=0$ if

$2x^2-4x+1=0$ . Via the quadratic equation we get

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

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

Via a sign chart (for instance with

$y'(0)=1$ ,

$y'(1)=-e$ and

$y'(2)=e^4$ ) we find that

$x=1-\frac{1}{2}\sqrt{2}$ is a maximum location and

$x=1+\frac{1}{2}\sqrt{2}$ is a minimum location.

Then

$y(1-\frac{1}{2}\sqrt{2})=(-1-\frac{1}{2}\sqrt{2})e^{1\frac{1}{2}-\sqrt{2}}$ is a maximum, and

$y(1+\frac{1}{2}\sqrt{2})=(-1+\frac{1}{2}\sqrt{2})e^{1\frac{1}{2}+\sqrt{2}}$ is a minimum.

Go on.

Antwoord 2 feedback

Wrong: Use the chain rule when differentiating.

See Chain rule.

Antwoord 3 feedback

Wrong:

$e^{x^2}+2x(x-2)e^{x^2} \neq 2x(x-2)$ .

Try again.

Antwoord 4 feedback

Wrong:

$e^{x^2}+2x(x-2)e^{x^2} \neq 2x(x-1)e^{x^2}$ .

Try again.