Probem52+Deborah,

Here we are looking to maximize the slope of the tangent line (which I think is the second derivative of the given function). We are given the curve y = 1 + 40x^3 - 3x^5. Since we are asked to find the points with maximum slope of the //tangent// line, we must take the derivative twice - once to figure out the equation of the tangent line and once after that to solve for the maximum. First, we'll take the derivative of the original curve, which results in the equation y' = 120x^2 - 15x^4. The maximum (or at least a maximum) of that will be found at the point where y'' = 0, meaning that the derivative function has hit a peak or a valley at a point of horizontal tangency (where its slope/derivative = 0). So we must take the derivative again, giving us y'' = 240x - 60x^3 = 0. Factoring tells us that 60x(4-x^2) = 0, or 60x[(2-x)(2-x)] = 0. Using the zero product property, x = 0 or 2.   Carolyn: End behavior: lim y' as x--> negative infinity= negative infinity lim y' as x--> infinity= negative infinity

y'(0) = 0 y'(2)= 240

Therefore the maximum is at x=2

//(ms. sweeney): very good!//