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Linearize $f$ near $x_{0}$.$$f(x)=\sqrt[3]{1+x}, \quad x_{0}=0.$$

$$y=\frac{1}{3} x+1$$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 6

Linearization and Differentials

Derivatives

Missouri State University

Harvey Mudd College

University of Michigan - Ann Arbor

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.

03:45

Linearize $f$ near $x_{0}$…

03:12

01:05

$f(x)=\sqrt{x}, \quad x_{0…

00:40

Use the given function $f$…

10:06

Let $f(x)=x \sqrt{x}, x \g…

05:25

$f(x)=\sqrt{2-x}+\sqrt{1+x…

01:03

$$\text { Let } f(x)=x \sq…

in this problem we are to linearize the function F of x is equal to the cube root of one plus X. Now we know that this is also the same as one plus X to the power of one third. To linearize the function, we used the equation of the linearize expression which is F. Of a plus F prime of a. Multiply by X minus A. Where A is given as X not. Is he going to 0? Now, first we find F prime of X which is we differentiate F of X. It gives us 1/3 Bracket one Plus X. Mhm. A play by one mm hmm Which is equal to 1/3. 1 plus. Thanks. Is our ex prime. Um To now we substitute our point F of zero. F prime of zero becomes if we have a zero here We get 1/3 F of zero. If put zero wherever there is an ex like here we get F zero equal to one because the cube root of one is still one. Now substituting into our L. Of X expression formula we get F of A or F zero as one plus F prime of a which is F prime of zero is one third. Well proper X zero. And our final answer becomes one plus X over three is our L. Of X. Yeah. Mhm

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