Answer
The tangent line has the largest slope at $~~x=-2~~$ and $~~x=2$
Work Step by Step
$y = 1+40x^3-3x^5$
$y' = 120x^2-15x^4$
Since $y'$ is the slope of $y$ at each point $x$, we can find the largest slope when $y'' = 0$:
$y'' = 240x-60x^3 = 0$
$x(240-60x^2) = 0$
$x = 0$ or $240-60x^2 = 0$
$x = 0$ or $x^2-4 = 0$
$x =-2, 0, 2$
When $x = -2$, then $y' = 120(-2)^2-15(-2)^4 = 240$
When $x = 0$, then $y' = 120(0)^2-15(0)^4 = 0$
When $x = 2$, then $y' = 120(2)^2-15(2)^4 = 240$
Note that the slope of the tangent line at any point $x$ is equal to the slope of the graph at the point $x$
The tangent line has the largest slope at $~~x=-2~~$ and $~~x=2$
