# B the graph is concave up for u1d465 2 and concave

• Homework Help
• 75
• 100% (1) 1 out of 1 people found this document helpful

This preview shows page 55 - 57 out of 75 pages.

(b) The graph is concave up for u1D465 < 2 and concave down for u1D465 > 2 , and the derivative is undefined at u1D465 = 2 . This is the case if the graph is vertical at u1D465 = 2 . One possible graph is shown in Figure 2.84. 2 u1D465 u1D453 ( u1D465 ) Figure 2.83 2 u1D465 u1D453 ( u1D465 ) Figure 2.84 12. We want to look at lim 0 ( 2 + 0 . 0001) 1∕2 − (0 . 0001) 1∕2 . As 0 from positive or negative numbers, the difference quotient approaches 0. (Try evaluating it for = 0 . 001 , 0.0001, etc.) So it appears there is a derivative at u1D465 = 0 and that this derivative is zero. How can this be if u1D453 has a corner at u1D465 = 0 ? The answer lies in the fact that what appears to be a corner is in fact smooth—when you zoom in, the graph of u1D453 looks like a straight line with slope 0! See Figure 2.85. −2 −1 0 1 2 1 2 u1D465 u1D453 ( u1D465 ) −0 . 2 −0 . 1 0 0 . 1 0 . 2 0 . 1 0 . 2 u1D465 u1D453 ( u1D465 ) Figure 2.85 : Close-ups of u1D453 ( u1D465 ) = ( u1D465 2 + 0 . 0001) 1∕2 showing differentiability at u1D465 = 0 13. (a) u1D445 u1D43Au1D440 u1D445 2 u1D45F u1D454 Figure 2.86 (b) The graph certainly looks continuous. The only point in question is u1D45F = u1D445 . Using the second formula with u1D45F = u1D445 gives u1D454 = u1D43Au1D440 u1D445 2 . Then, using the first formula with u1D45F approaching u1D445 from below, we see that as we get close to the surface of the earth u1D454 u1D43Au1D440u1D445 u1D445 3 = u1D43Au1D440 u1D445 2 . Since we get the same value for u1D454 from both formulas, u1D454 is continuous. (c) For u1D45F < u1D445 , the graph of u1D454 is a line with a positive slope of u1D43Au1D440 u1D445 3 . For u1D45F > u1D445 , the graph of u1D454 looks like 1∕ u1D465 2 , and so has a negative slope. Therefore the graph has a “corner” at u1D45F = u1D445 and so is not differentiable there.
202 Chapter Two /SOLUTIONS 14. (a) The graph of u1D444 against u1D461 does not have a break at u1D461 = 0 , so u1D444 appears to be continuous at u1D461 = 0 . See Figure 2.87. −2 −1 1 2 1 u1D461 u1D444 Figure 2.87 (b) The slope u1D451u1D444 u1D451u1D461 is zero for u1D461 < 0 , and negative for all u1D461 > 0 . At u1D461 = 0 , there appears to be a corner, which does not disappear as you zoom in, suggesting that u1D43C is defined for all times u1D461 except u1D461 = 0 . 15. (a) Notice that u1D435 is a linear function of u1D45F for u1D45F u1D45F 0 and a reciprocal for u1D45F > u1D45F 0 . The constant u1D435 0 is the value of u1D435 at u1D45F = u1D45F 0 and the maximum value of u1D435 . See Figure 2.88. u1D45F 0 u1D435 0 u1D45F u1D435 Figure 2.88 (b) u1D435 is continuous at u1D45F = u1D45F 0 because there is no break in the graph there. Using the formula for u1D435 , we have lim u1D45F u1D45F 0 u1D435 = u1D45F 0 u1D45F 0 u1D435 0 = u1D435 0 and lim u1D45F u1D45F + 0 u1D435 = u1D45F 0 u1D45F 0 u1D435 0 = u1D435 0 . (c) The function u1D435 is not differentiable at u1D45F = u1D45F 0 because the graph has a corner there. The slope is positive for u1D45F < u1D45F 0 and the slope is negative for u1D45F > u1D45F 0 . 16. (a) Since lim u1D45F u1D45F 0 u1D438 = u1D458u1D45F 0 and lim u1D45F u1D45F + 0 u1D438 = u1D458u1D45F 2 0 u1D45F 0 = u1D458u1D45F 0 and u1D438 ( u1D45F 0 ) = u1D458u1D45F 0 , we see that u1D438 is continuous at u1D45F 0 .