Show that the tangent line to the curve y = (x^2 + x - 2)^3 + 3 at (1, 3) is also tangent to the curve at another point.
Like my previous question I got the derivative
being 3(2x+1)(x^2 +x -2)^2
I am not sure where to go from here.
Thank you!
Show that the tangent line to the curve y = (x^2 + x - 2)^3 + 3 at (1, 3) is also tangent to the curve at another point.
Like my previous question I got the derivative
being 3(2x+1)(x^2 +x -2)^2
I am not sure where to go from here.
Thank you!
To find the other point that the tangent line touches, note that
$\displaystyle \frac{dy}{dx}=0 \text{ and also, your tangent line is given by } y=3, \text{ so set } y=3 \implies 3=(x^2+x-2)^3 + 3 \implies 0=(x^2+x-2)^3 \implies 0=(x^2+x-2) \implies \cdots \text{ (you can practice your factoring here! =) )}$
Just a note, in this question the tangent line touches 2 local extrema at the same level (i.e. y=3), but not necessarily.