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Use the first and second derivatives to sketch the graph of the given equation. Also include the intercepts, whenever they are easily determined.$$f(x)=x^{3}+x+2$$

Additional $x$ -intercepts at $\pm 2 \sqrt{3}$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 3

Concavity and the Second Derivative

Derivatives

University of Michigan - Ann Arbor

University of Nottingham

Boston College

Lectures

04:40

In mathematics, a derivative is a measure of how a function changes as its input changes. Loosely speaking, a derivative can be thought of as how much one quantity is changing in response to changes in some other quantity; for example, the derivative of the position of a moving object with respect to time is the object's velocity. The concept of a derivative developed as a way to measure the steepness of a curve; the concept was ultimately generalized and now "derivative" is often used to refer to the relationship between two variables, independent and dependent, and to various related notions, such as the differential.

30:01

In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value (the rate of change of the value of the function). If the derivative of a function at a chosen input value equals a constant value, the function is said to be a constant function. In this case the derivative itself is the constant of the function, and is called the constant of integration.

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Use the first and second d…

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question 48 would like you to use the 1st and 2nd derivatives to sketch the graph of f of x equals X cubed plus X plus two um, and also including intercepts if they're easy to find. So starting off, we can find critical points by taking a prime of X is equal to three x squared plus one. Ah, set that equal to zero. Ah, if you saw for ex you, though, get X is equal to plus or minus square root of negative one third, which can't have because it's an imaginary number. Therefore, there aren't any critical points. We can then look for common cavity and inflection points with the double prime of X, which is equal to six x. So if F double prime of X is equal to zero, there is an inflection point, which means that at X equals zero, there is an inflection point, which is half of zero is equal to just to see. Our inflection point is at 02 From there we can look for any kind of intercepts or additional points that are helpful. So, uh, plugging in or setting this I'm sorry. Setting this equals zero is equal to execute plus X plus two. Uh, this becomes zero when X is equal to negative one. Therefore, we have the point negative 10 and then just testing points that are before and after this inflection point here. So f double prime of two is going to be 12, which means it's concave up after our inflection point and f double prime of negative two is negative. 12. So we're concave down before that inflection point and you can also find just half of two is equal to 12 and f of negative two is equal to negative eight. Now, just going ahead and plotting these points, we have negative one zero there and you have two all the way up to 12. And you have two over here at about negative eight. Therefore, knowing that this side is concave up, we can draw it like this and this would be concave down, so it should look like that. And that is your answer to question 48

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