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a) sketch the graph of the given function, and then draw the tangent line at the point $P$. (b) Using your sketch, approximate the slope of the curve at $P,$ (c) Use (1) to determine the exact value of the slope at $P$.$$f(x)=x^{3} \quad P(2,8)$$

A. (2,8)C. 12

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 1

Slope of a Curve

Derivatives

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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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a) sketch the graph of the…

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here were given a function equal to execute and asked to find its slope at 0.28 I've gone ahead and sketched it for us. It looks something like this. 0.28 sits up here. We want to find the slope of this curve at 28 To do that, we could draw a yes tangent line to it. Here, we need to find another point somewhere along it. So let's choose a point really close to this and plug it into our function. Let's use X is equal to 2.1 plugging that in. We have 2.1 cube will be equal to our Y value and 2.1 cute is equal to 8.12 That gives us our Y value. Now, if we were to plug in our new X and Y values, we just found along the ones that were given to us into what we know to be a slope function where M is equal to y tu minus y one all of our x two minus x one. We can plug these values in to give us 8.1 to minus minus. Aged all over 2.1 minus two. That will give us a value equal to 0.12 all over 0.1 Dividing these right here. We end up getting a slope equal to 12.6 So we can assume to have a slope equal to 12. Go ahead and say that we have a slope at 0.28 is equal to 12.

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