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(a) Use Newton's method to find the critical numbers of the function $$ f(x) = x^+ - x^4 + 3x^3 - 2x $$ correct to six decimal places.(b) Find the absolute minimum value of $ f $ correct to four decimal places.
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Calculus 1 / AB
Calculus 2 / BC
Chapter 4
Applications of Differentiation
Section 8
Newton's Method
Derivatives
Differentiation
Volume
Missouri State University
Campbell University
Oregon State University
University of Nottingham
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(a) Use Newton's meth…
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a. Locate the critical poi…
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question 35 way Have bets X to the six x to the fore. The three execute minus two x So are right of X is going to equal six X minus four x Q waas Nynex. Where mine too? Looking at the graph, we have zeros at negative 1.3 negative one for and playing five. I'm gonna use my devil after all 30 x to the fore minus one X squared plus 18 x no, I'm gonna say except to equals except one minus my I'm trying over my f ever prime. So when I do that Wait, um exit to of negative 1.3 negative 1.3 is my except one All right or my ex too. I get negative one point 29 3344 And for my ex of three, which will equal my exit or I get approximately negative White too. Nine, 3227 When I plugged in the negative 0.4 for my ex of two. I get negative. Wait. Or or 3755 The three then is a approximately negative or who were 173 and except for and exit five Well negative. Only four were 1731 So the equally gentry stop. And unless I have a 0.5 camels room negative except two equaled point 507 nine 37 except three. Then equals X a poor which equals quite 507854 Now her B is asking us, Um yeah, absolute minimum. So I'm looking at my critical numbers in my f prime and F problem has two critical numbers going from a decreasing toe on increasing or a negative Quite positive. So I'm going to look at my, uh, in my function in my function act negative 1.29 which is one of my critical numbers. 3227 And when I put that into my function, I get it negative. 2.2 12 And when I look at other critical number 0.507 854 I get negative. Quite 67 to 1, for that is my absolute minimum
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