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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
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