Exercise 2 | Wiskunde op Tilburg University

Determine all the extrema of

$y(x)=x^3+5x^2-7$ for

$x\geq -1$ .

Antwoord 1 correct

Correct

Antwoord 2 optie

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

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.

Antwoord 2 feedback

Wrong: Consider the domain of the function.

Try again.

Antwoord 3 feedback

Wrong: Consider the boundary points.

See First-order condition extremum.

Antwoord 4 feedback

Wrong:

$y'(x)\neq 3x^2+10-7$

See Derivative.