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Discuss (i) the continuity and; (ii) the differentiability of the given function. (iii) If a discontinuity is removable, redefine the function so as to make it continuous there. (iv) Is it now differentiable?$$t(s)-\sqrt{1-s^{2}}$$

Continuous and diff. if -16 s 61

Calculus 1 / AB

Chapter 2

An Introduction to Calculus

Section 3

Limits and Continuity

Derivatives

Missouri State University

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.

02:01

Discuss the continuity of …

03:18

Let $f(x)=|x-1|([x]-\{x\})…

here, we're looking at the continuity of this function on the closed interval from negative 323 The only thing that's going to restrict the continuity of this function is the value of the argument of the square root function. So we need nine minus T squared to be greater than or equal to zero because we can't take the square root of a negative number. Now this function, if we factor it, we can factor this as a difference of two squares and then if we do a sign analysis of this function, so it has two zeros at negative three and positive three, we can test the sign of each factor within the intervals. So we have three intervals. So we put in a test number in each interval, We can check the sign. So if we check the first interval, let's take a test number like -4. So 3 -1 is going to end up to be a positive number. In that case, if we put a negative for for T and three plus T will be a negative number. In that case, if we move on to the next interview we can put in zero as a test number because that's between negative three and three. That makes this factor positive, That makes this factor positive as well. And then if we go to the last interval we could put in four as a test number, so three minus T becomes negative and three plus T would be positive and then we can take the product of those factors. So in the first of all, we have a positive times a negative which is a negative. In the second interval we have a positive times a positive which is a positive. And in the third interview we have a negative times a positive which is negative. Now note that this is always greater than the recorder zero in the interval between negative three and three. So this function is continuous for all values that are in that interval, so it's continuous for negative three 23

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