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Let $f(x)$ be defined on the closed interval [0,1] by the rule:$$f(x)=\left\{\begin{array}{cl}2 x^{2} & \text { if } 0<x<1 \\1 & \text { if } x=0 \text { or } x=1\end{array}\right.$$(a) Does fhave extrema on [0,1]$?$ (b) Is this a violation of Theorem $1 ?$

(a) No(b) No

Calculus 1 / AB

Chapter 3

Applications of the Derivative

Section 1

Extrema of a Function

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.

01:56

Let $f(x)=x^{-1 / 2}$ on t…

00:44

A minimum with no derivati…

01:13

Yeah. First we want to find the critical point of this function so we take it to relatives. Two, two X square. Richard calls for acts and if allowed to drive it. How? Because there are all it will give us access. It was terrible. And next we want to find the extreme during the interval zero to work know that when X equals zero or one. Its function value is one according to this and well X approaches zero, it's functional value approaches zero and X approaches wild function value approaches to. Therefore this function has neither mean maximum not meaning. So does this example violate theory? One. The answer is no. Syrian one says that if a function defined by Y equals effects is continuous on the closed bounded in general, then it obtains both a maximum and the minimum value on this interval. But not that here. This function is not continuous between the interval 0 to 1. The function is not continuous at their it not a continuous one either enhance. This function can approaches zero approaches to infinity, but it cannot reach. That is the reason why this function has neither maximum not minimum.

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