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.
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.
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).
Besides the chain rule you should also use the Productrule (film).
Antwoord 4 feedback