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p 278 # 69: Eliana

Interval = [0,30] degrees C V = 999.87 - 0.06426T + 0.0085043T^2 - 0.0000679T^3 V` = -0.06426 + 0.0170086T - 0.0002037T^2 I plugged this into the quadratic formula on my calculator, the value I got that worked within the given interval was 3.9665 degrees C

What is the volume of water at this value of C? And what about the value of V at the endpoints? How does this compare to your value V(3.9665)?

V(0) = 999.87 V(3.9665) = 999.74 --> this makes this critical point the local and absolute minimum of the interval, since values on either side of it are higher, and it is the lowest point on the interval as a whole. V(30) = 1003 --> Absolute max

Remember that the procedure is to find the value of the function at the endpoints, the value of the function at each critical value, and then one (or more) of these will be your absolute extrema! (ms. sweeney)

p. 278 # 71: Sophia

Interval = [0,30] (August 1993 - August 2003) S(t) = -0.00003237t^5 + 0.0009037t^4 - 0.008956t^3 + 0.03629t^2 - 0.04458 + 0.4074 Minimum = .436 @ t=4.618 Maximum = .391 @ t=.855 The min and max values of t show that sugar was most expensive in March 1998 and cheapest in June 1994.

Did you check the value of the function at the endpoints? You should show this. Also, how did you arrive at these values? What is your derivative? Show how you set it equal to zero, and explain how you used the derivative to come up with these critical values! (ms. Sweeney)


 * 73: Jenna

a. V(r)=k(ro-r)r^2 [1/2ro, ro) V(1/2ro)=k(1/2ro)(1/4ro)=1/8kro V(ro)=0 V(r)=kro(r^2)-kr^3 V'(r)=2kro(r)-3kr^2 2kro(r)-3kr^2=0 r=2/3ro

b. V(2/3ro)=k(1/3ro)(4/9ro^2) V=4/27k(ro)^3 And so what is your conclusion? Which actually yields the maximum velocity? Give an answer! (ms sweeney)

5 = 2 + (x-5)^3 3 = (x-5)^3 ?????? The question asks to show that 5 is a critical value, so by definition of critical value we need to start with taking the derivative, and showing that if you set the derivative equal to zero (or look for where it is undefined), you will end up with x = 5. What you are doing is not really what is asked for; finding the value of x for which the output is 5.
 * 74: Shane
 * a)** g(x) = 2 + (x-5)^3

5 = 3(x^2 - 10x + 25) 5/3 = x^2 - 10x + 25 0 =x^2 - 10x + 23.33 __QUAD: x = 6.29228, 3.70771__ f(6.29228) = 4.158 f(3.70771) = -0.158
 * b)** g'(x) = 3(x-5)^2

Again, remember that the definition of a critical value is that it is the location where the derivative is equal to zero; so set g'(x) = 0 and solve for x!!

Prove that this function does not have a local minimum or maximum--> f(x) = x^101 + x^51 + x + 1 - First we take the derivative, so: f'(x)= 101x^100+51x^50+1 - Then set it equal to 0, to see where the local min and max are--> 101x^100+51x^50+1=0 - Make u equal to x^50, so 101u^2+51u+1=0 - Solve using the quadratic formula, so u = [-51 ± 13√(13)] / 202 - Plug back in x^50 for u, so x^50=[-51 ± 13√(13)] / 202 - Take the 50th root of each side x = [(-51 ± 13√(13)) / 202]^(1/50) x ≈ (-0.0204348189)^(1/50) and (-0.484515676)^(1/50) You can not take the root of a negative number, therefore there are no solutions to where the derivative of f(x) = x^101 + x^51 + x + 1 equals 0. Since that is the case, there are no local minimums or maximums
 * 75: Julia