Exercise 6 | Wiskunde op Tilburg University

Determine all the values of

$x$ such that the tangent line to the graph of the function

$y(x)=(x-1)(x^2-1)^3$ is horizontal.

Antwoord 1 correct

Correct

Antwoord 2 optie

$x=-1$ ,

$x=-\frac{1}{7}$ ,

$x=\frac{1}{7}$ ,

$x=1$

Antwoord 2 correct

Fout

Antwoord 3 optie

$x=-1$ ,

$x=0$ ,

$x=1$

Antwoord 3 correct

Fout

Antwoord 4 optie

$x=1$

Antwoord 4 correct

Fout

Antwoord 1 optie

$x=-1$ ,

$x=-\frac{1}{7}$ ,

$x=1$

Antwoord 1 feedback

Correct:

$y'(x)=(x^2-1)^3+(x-1)3(x^2-1)^2\cdot 2x=(x^2-1)^2\Big((x^2-1)+6x(x-1)\Big)$ .

The tangent line is horizontal if

$y'(x)=0$ . Hence,

$x^2-1=0$ or

$(x^2-1)+6x(x-1)=0$ .

$x^2-1=0$ gives

$x=-1$ or

$x=1$ .

$(x^2-1)+6x(x-1)=7x^2-6x-1$ and hence, we use the quadratic equation:

$x_1=\dfrac{6-\sqrt{(-6)^2-4\cdot7\cdot(-1)}}{2\cdot 14}=-\frac{1}{7}$ and

$x_2=\dfrac{6+\sqrt{(-6)^2-4\cdot7\cdot(-1)}}{2\cdot 14}=1$ .

Hence,

$x=-1$ ,

$x=-\frac{1}{7}$ ,

$x=1$ .

Go on.

Antwoord 2 feedback

Wrong:

$x=\frac{1}{7}$ is not a solution of

$(x^2-1)+6x(x-1)=0$ .

Try again.

Antwoord 3 feedback

Wrong:

$y'(x)\neq 3(x-1)(x^2-1)^2\cdot 2x$ .

Besides the chain rule you should also use the Productrule (film).

Antwoord 4 feedback

Wrong: The tangent line is not horizontal for

$f'(0)$ .

See Example 2 (film)