Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Discover in-depth answers to your questions from a wide network of experts on our user-friendly Q&A platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
The points on the curve y=1+[tex]60x^{3}-2x^{5}[/tex] have the largest slope are
(-3, -1133) and (3, 1133).
Given the equation of curve be y = 1+[tex]60x^{3}-2x^{5}[/tex]
We are required to find the points at which the slope of tangent is large.
The derivative provides us the slope at any given point, so
y’ = [tex]180x^{2} -10x^{4}[/tex]
Now, in order to maximize this, we need to take the derivative again and put it equal to zero:
y” = 360x – [tex]40x^{3}[/tex]= 0
Solve to find the value of x to get x = -3, 0, or 3, these are our critical points.
Use these back into your slope equation which is y=1+[tex]60x^{3}-2x^{5}[/tex]
y'(-3) = [tex]180(-3)^{2} -10(-3)^{4}[/tex]=810
y'(0) = =[tex]180(0)^{2} -10(0)^{4}[/tex]0
y'(3) = [tex]180(3)^{2} -10(3)^{4}[/tex]=810
So your curve has the greatest slope at x = -3 and 3
Use this into your original equation for
y(-3) =[tex]1+60(-3)^{3} -2(-3)^{5}[/tex]= -1133
y(3) = [tex]1+60(3)^{3} -2(3)^{5}[/tex]=1133
So our curve has the greatest slope at points (-3, -1133) and (3, 1133).
Hence the points on the curve y=1+[tex]60x^{3}-2x^{5}[/tex] have the largest slope are
(-3, -1133) and (3, 1133).
Learn more about differentiation at https://brainly.com/question/954654
#SPJ4
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.
