By W. Chen

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1. Consider the the function function ff (x) (x) = = eexx2 −2x Example .. The graph does not intersect the x-axis, and intersects the y-axis at the point (0, 1). Also, f (x) → +∞ as x → +∞ and as x → −∞. On the other hand, it follows from the Chain rule that 2 2 2 −2x f (x) = (2x − 2)exx −2x <0 =0 >0 if x < 1, if x = 1, if x > 1. Hence there is a stationary point at x = 1. Also the function is decreasing when x < 1 and increasing 2 2 2 −2x when x > 1. Now f (x) = ((2x − 2)22 + 2)exx −2x > 0 always.

A) Find the first derivative dx b) Find the coordinates of the point(s) on the curve where the tangent(s) is (are) horizontal. c) Find the second derivative d2 y at the point (−2, 1). dx2 14. For each of the following, find the first derivative y and second derivative y in terms of x and y by implicit differentiation: √ √ a) x3 + y 3 = 1 b) x + y = 1 c) x2 + 6xy + y 2 = 8 15. The function y = f (x) is defined implicitly by sin x + y 3 = 8. Find 16. The function y = f (x) is given implicitly by x3 y + y 3 = 9.

8. If we consider the function f (x) = x3 − 3x − 1, then finding where the graph intersects the x-axis becomes an extremely difficult problem. Of course, it is not absolutely crucial to locate all the points where the graph of the function intersects the coordinate axes. STEP 5. ASYMPTOTES. Consider first an example. 9. Try to draw a rough sketch of the graph y = 1/x. It is easy to see that the graph gets rather close to the coordinate axes. Try to draw next a rough sketch of the graph y = 1 + 1/x.