## Some answers to the review problems for the second exam in Math 151:04-06, fall 2003

AB Although students will work on these review problems on Thursday, I invite e-mail with solutions. Each student first should try the problem assigned to their last name. For example, Jane or John Smith should send me e-mail with the answer to #10. I will post the answers here, and change the red background to green at that time. Students giving answers will receive my thanks and will be identified here, additionally obtaining notoriety, glory, etc.

Michael Pandolfo writes:
For part a I got A(1) = 3, ... A'(x) = (23-7(5x^2-3))^(1/2)*(10x) and A'(1) = 30.
b. y-3 = 30(x-1)
c. A(.95) = 1.5
d. Yes A"(1) = -(260/3) The value found in part c is larger than the actual value of A(.95) because A"(x) < 0 so the curve is concave down and the tangent line is above the curve.

Sherman Yang writes:
a. If you input infinity for the top and the bottom, it will be infinity over infinity so you take the l'hopitals. then input infinity to the top and bottom again and it will become 1/infinity which is 0.
b. Just plug in 3 and it is 0/6 so the answer is 0.
c. Separate the equation into two so it will be sinx/(e^x-1)^2 - x/(e^x-1)^2. Since the range of sinx is -1 to 1, you plug in infinity for -1/(e^x-1)^2 and 1/(e^x-1)^2. Both of them will equal 0 so 0