Can anyone verify if I am doing the following correct? One of the practice problems in my book says to Find equations for the horizontal tangents to the curve $y=x^3-3x-2$.
I have the following worked out via derivatives:
$y=x^3-3x-2$
$y^1=3x^2-3$
$0=3x^2-3$
$3=3x^2$
$3/3 = x^2$
$1=x^2$
$sqrt(1)=sqrt(x^2)$
$1=x$
I then plugged 1 inf or x in my original equation:
$y=(1)^3-3(1)-2$
$y=1-3-2$
$y=-4$
There is a horizontal tangent at Point(1,-4). Then based on my graph for $y=x^3-3x-2$, I believe there is also a horizontal tangent at Point(-1,0) along the origin? Thus I plugged in 0 for x in my original equation as well:
$y=(0)^3-3(0)-2$
$y=0-0-2$
$y=-2$
This then would be Point (0,-2), but that is not a horizontal tangent on my graph?
Best Answer
The first tangent is correct. For the second tangent, note that $x^2=1$ has another root at $x=-1$.
Alternatively, note that $$y=x^3-3x-2=(x+1)^2(x-2)$$ This means there is a double root at $x=-1$, so the x-axis itself ($y=0$) is a horizontal tangent.