Problem

Find the critical numbers of the function. $ f…

06:56

Problem 29 Easy Difficulty

Find the critical numbers of the function.

$ f(x) = 4 + \frac{1}{3}x - \frac{1}{2}x^2 $

Answer

$$f(x)=4+\frac{1}{3} x-\frac{1}{2} x^{2} \Rightarrow f^{\prime}(x)=\frac{1}{3}-x, f^{\prime}(x)=0 \Rightarrow x=\frac{1}{3}$$

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Video Transcript

who would like to find? The critical numbers of the function F. Of X equal four plus one third X minus one half X square. Not that the critical numbers of the function. Our numbers into the main of these functions such that Mhm. Such that the derivative at that point doesn't exist or is zero. So that's it. That's his finishing. And first of all, the domain of the given function is the real numbers. I'm talking now about this function. That is because it is it's a polynomial of second degree. So is well defined at every real number eggs. And so the domain of the function of the whole real numbers. Mhm. Now let's see what is the derivative of F at any point is one third minus X. Because his negative 1/2 times two Times X. to the one. So this is a derivative is defined at every point in the real numbers. So the derivative exists for any real number. That is for any X. In the domain of the function. So this part do not apply in this case, So we only get five points for the theory of a TV- zero. Yeah. So the derivative of this function is zero if and only if one third minus six is zero, which is equivalent to X Equal 1 3rd, and that's the only value for which the derivative zero. And taking into account that it exists everywhere, it is well defined and exists at every real number. I'm talking about the relative, then this is the only critical point to conduct critical number of fx equal four plus one third of x minus one half X. Square. Yes, X Equal 1 3rd. Mhm. So this is you find out the answer.

Oswaldo J.

Universidad Central de Venezuela

Topics

Derivatives

Differentiation

Volume

Calculus: Early Transcendentals

Chapter 4

Applications of Differentiation

Section 1

Maximum and Minimum Values

Problem