Determine all the extrema of $y(x)=x^3+5x^2-7$ for $x\geq -1$.
Antwoord 1 correct
Correct
Antwoord 2 optie
- $y(-3\frac{1}{3})=11\frac{14}{27} $ is a maximum
- $y(0)=-7$ is a minimum
Antwoord 2 correct
Fout
Antwoord 3 optie
$y(0)=-7$ is a minimum
Antwoord 3 correct
Fout
Antwoord 4 optie
$y(-1\frac{2}{3}+\frac{1}{6}\sqrt{184})=-5\frac{1}{40}$ is a minimum
Antwoord 4 correct
Fout
Antwoord 1 optie
- $y(-1)=-3$ is a boundary maximum
- $y(0)=-7$ is a minimum
Antwoord 1 feedback
Correct: $y'(x)=3x^2+10x$. The zeros of $y'(x)$ are therefore, $x=0$ and $x=-3\frac{1}{3}$, but since $x=-3\frac{1}{3}$ is outside the domain of the function only $x=0$ remains. Via a sign chart (with for instance $y'(-\frac{1}{2})=-4\frac{1}{4}$ and $y'(1)=19$) we find that $x=-1$ is a maximum location and $x=0$ is a minimum location.
Hence, $y(-1)=3$ is a boundary maximum and $y(0)=-7$ is a minimum.
Go on.
Hence, $y(-1)=3$ is a boundary maximum and $y(0)=-7$ is a minimum.
Go on.
Antwoord 2 feedback
Wrong: Consider the domain of the function.
Try again.
Try again.
Antwoord 3 feedback
Antwoord 4 feedback