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Use the first derivative to determine where the given function is increasing and decreasing.$j(x)=\sqrt{8-2 x}$

Dec on its domain $(-\infty, 4)$

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 2

The First Derivative Test

Derivatives

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Campbell University

Idaho State University

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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find the derivative of J. F. X. And we're going to do this by going to rewrite it slightly and have it be to the one half power, which is the same thing as the square root. And so we're going to use chain rule to bring down the exponents. Keep the inside expression And subtract one from the exponent times the derivative of the inside function, which is -2. We want this to equal zero. So to get what X value is equal zero. We want to solve, see you what X value will make this zero. And so since we have a product here, we just have to look at the inside since native to doesn't have a variable. So at the inside let's look at where how we can make this inside function equals zero. So the whole expression equal zero. And we get that at x equals four. So keep in mind that in the original equation we have a square root, you cannot have a rational number G a square root of a negative number. So with that in mind let's pick a value slightly below and slightly above. I'll do three and five. So at five if we were to plug in five to this X value, we have a negative number and a negative square root of a negative number is not a rational number. So the function doesn't actually exist past the value of four. And then at three if we were to plug that in into our X, we get negative. So based on this, our function is decreasing when X is less than four and it does not exist or rationally doesn't exist has when X is greater than for.

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