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
- $y(0)=-2$ is a maximum
- $y(2)=0$ is a minimum
Antwoord 3 correct
Fout
Antwoord 4 optie
- $y(0)=-2$ is a maximum
- $y(1)=-e$ is a minimum
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.
$y'(x)=0$ 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
Antwoord 3 feedback
Wrong: $e^{x^2}+2x(x-2)e^{x^2} \neq 2x(x-2)$.
Try again.
Try again.
Antwoord 4 feedback
Wrong: $e^{x^2}+2x(x-2)e^{x^2} \neq 2x(x-1)e^{x^2}$.
Try again.
Try again.