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Find the critical numbers of the function.
$ f(x) = 2x^3 + x^2 + 2x $
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01:32
Fahad Paryani
00:34
Amrita Bhasin
01:31
Chris Trentman
Calculus 1 / AB
Calculus 2 / BC
Chapter 4
Applications of Differentiation
Section 1
Maximum and Minimum Values
Derivatives
Differentiation
Volume
Missouri State University
Baylor University
University of Michigan - Ann Arbor
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Find the critical numbers …
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Find any critical numbers …
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Yeah. We want to find the critical numbers of the function F of X equal to x cubed plus X squared plus two X. So, first we find the derivative of F. That function is six X squared plus two eggs plus two. And this is relative exist at any ex any real number eggs. The main in fact of the function is the whole real numbers. Because this expression is paranormal. Is it fine for every real number X. So the men of F. S are the real numbers and in a relative exists for every eggs in the real numbers. That is is two facts together implies that the only critical points of function F Are those eggs for which there are TV zero. Mhm. Yeah. That is because remember the definition of critical number of a function. Yeah. All right. The X. Or values in the domain of the function for which the relative does not exist or It exists and is equal to zero. So uh that's why we look first at the domain, the function that is the real numbers. And then we see where derivative exists. In this case it exists for every real number. That is for every X. In the domain of the function. Therefore the only critical points or numbers of these functions are those values of X. In the real numbers for which the derivative is equal to see. Mhm. So the equation we gotta solve is this f derivative equal zero. And that's equivalent to looking at the formula here, derivative six X square last two eggs Last two equals 0. And that's the same thing in two common factory at two times three X square plus eggs plus one. Because two is against a constant different from zero is equivalent to, sorry, here is equal zero. And is going to that the expression inside parenthesis through three X squared plus X plus one Be equal to zero. That that is the equation we get us solved with this one here because sold in this equation we have all the eggs for which the first derivative of the function is equal to zero. So let's find that by applying the formula. Or best, we're going to calculate the discriminate of this equation because second degree polynomial equals zero. So you see it's criminal expression which is the square minus four K. C or B is a division of X to the one a suggestion of X square and sees an independent term. So in this case is equal to one square Wayne is four times 3 times one That is one 12. That is -10. And because this expression is going to appear inside the square root in the general formula, dissolution of the second order and a question, you know that with this condition here, we won't have real solutions, then, um There is no number X. And the real numbers such that yeah, three X square plus six plus one equals zero. That is the same as saying that the first derivative of F Different from zero for all eggs, Nah, real numbers. And then taking into account what we said above, we can conclude that this function has no critical points and that the final answer cause this problem.
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