Asked by Suzy
What is the smallest possible slope for a tangent to y=x^3 - 3x^2 + 5x?
(I'm unsure how to approach this problem, if you know how to solve it, please explain step by step. THANK YOU!!!)
(I'm unsure how to approach this problem, if you know how to solve it, please explain step by step. THANK YOU!!!)
Answers
Answered by
Damon
dy/dx = slope = 3 x^2 - 6 x + 5
where is that max or min ?
where its derivative is 0
d^2y/dx^2 = 0 = 6 x - 6
so x = 1
is that a max or a min?
take the ext derivative, if + it will be headed up and a min.
d^3y/dx^3 = 6
that is positive so the min slope is at x = 1
that slope is
3 x^2 - 6 x + 5 = 3-6+5 = 2
where is that max or min ?
where its derivative is 0
d^2y/dx^2 = 0 = 6 x - 6
so x = 1
is that a max or a min?
take the ext derivative, if + it will be headed up and a min.
d^3y/dx^3 = 6
that is positive so the min slope is at x = 1
that slope is
3 x^2 - 6 x + 5 = 3-6+5 = 2
Answered by
Suzy
THANK YOU SO MUCH!!!
Answered by
Damon
You are welcome.
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