By Dovermann K.

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**Example text**

To see this, square the equation and write it in the form x2 + y 2 = 1, which is the equation of the circle. Thus √ we are saying that the slope of the tangent √ line to the circle 2 at a point (x, 1 − x ) in the upper hemisphere is −x/ 1 − x2 . 4. What is the slope of the tangent line to the circle at a point (x, y)? 4: The radial line is perpendicular to the tangent line. 1 The slope of the radial line is y/x. In analytic geometry you (should have) learned that two lines intersect perpendicularly if the product of their slopes is −1.

Find the tangent line to the graph of ln x at the point (1, 0). By now you may have gotten the impression that all functions are differentiable. This is not so. 14. 3. 7: The absolute value function is not differentiable at x = 0. Solution: There is no potential tangent line which is close to the graph of f (x) near x = 0 in the sense in which it has been specified in the definition of differentiability. Zooming in on the point (0, 0) does not help, the picture remains the same. You can give an analytical argument.

Why? The program makes substantial round-off errors in the calculation. Which one is the correct graph? Calculus will CHAPTER 1. 18: p(x) = (x + 1)6 tell you that the second graph cannot have come close to the truth. Is the first one correct? This is difficult to tell, particularly, as y values are indistinguishable. The program shows 0’s at all ticks on this axis. True, the numbers are small, but they are certainly not zero. Still, the general shape of the graph in the first figure appears to be quite accurate.