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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.$$f(x)=5+x^{2 / 3}$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 3

Concavity and the Second Derivative

Derivatives

Missouri State University

Harvey Mudd College

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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Use the First Derivative T…

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Find the $x$ -intercepts o…

question 61 is asking you the 1st and 2nd derivative of F of X equals five plus X to the two thirds to graph this function and also include any intercepts if they're easy to find. So starting off s prime of X is equal to to over three X to the negative one third. This is just equivalent to 2/3 X to the one third. This will be undefined at X equals zero so we can treat this kind of like a critical point to test our con cavity. Uh, you also know that there is a y intercept, um, at X equals zero because zero is equal to five. So that's a point that you can use now. F double prime affects is equal to negative two over nine X to the negative four thirds. So negative two over nine X to the four thirds. Now we can test our con cavity. So when X is plus than zero testing F double prime of negative one is negative 18, which means we're concave down and f double prime of one. So when X is greater than zero is equal to negative 18 as well. So you're also concave down. So essentially what your graph is going to look like, uh, starting at the point five zero on the or 05 on the Y axis and you know it's concave down on either side. Therefore, your graph looks like this, and that's your answer to question 61.

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