#1
Nov903, 02:09 AM

P: n/a

Hey Everyone. I'm at my wits end on these problems, and was hoping for some help. Thanks for any input given.
1)Given G(t)=(1.35)^t+7 A)Describe the domain, range, intercept(s) and asymptotes of G(t). B)Write a formula for the inverse function. C)Describe the domain, range, intercept(s) and asymptotes of G^1(t). D)Use a table to verify that you computed G^1(t) correctly. E) Sketch both functions on the same set of axes. 2)Radioactive iodine is a byproduct of a certain type of nuclear reaction. Its half life is 60days. Suppose that an accident occurs and 45pounds of radioactive iodine is released into the environment. The amount of radioactive iodine decays according to the model: f(t)=ab^t. A)Write a formula for f(t). B)By what percent is the amount of radioactive iodine decreasing each day? C)Calculate and interpret the following quantities. Explain why the values are not equal. f(4)f(0)/4 f(7)f(3)/4 D)How long will it take until 80% of the released amount has decayed? 



#2
Nov903, 02:28 AM

P: 20

rather than complete the two four part homework problems for you, think of what you already know about the function. What is the general form for the function. what shifting is ocurring. how does this affect the overall function.
for domain this of what values for t won't yield a result. For Range consider what values for G(t) cannot come from the equation (hint:can a postive raised to any power be negative?) Intercepts consider when t is 0 and if/when G(t) is 0. 



#3
Nov903, 03:18 AM

P: 1,572

surely there are examples in the text like this, aren't there?
for c, you can use a general principle that the domain and range of f are the range and domain of f^{1}. 


#4
Nov903, 04:22 PM

P: n/a

Pre calc help(first post)
I'm having trouble figuring out the inverse of G(t)=(1.35)^t+7.
Also having trouble with putting problem 2A) into a formula. 



#5
Nov903, 07:22 PM

P: 640

Domain: all, range: all, no vertical asymptote, horizontal *** at y = 7. g^1(x) = log(t/7)/log(1.35)



#7
Nov903, 10:30 PM

P: n/a

So for the formula G(t)=(1.35)^t+7
Is the vert. intercept: (0,1.35) No horizontal intercept? 



#8
Nov903, 10:37 PM

P: 1,572

by the way, especially in light of the fact that at least one person here has given you wrong information, be sure to catch an explanation and/or proof of any claim made before you believe it.
the vertical intercept is (0,y) and the horizontal intercept is (x,0). if you let t=0, what do you get? ie, what is G(0)? it's not 1.35. if you let G(t)=0 what do you get for t? ie if =01.35^t+7, what is t? what you get, if anything (hint), is the horizontal intercept. 



#9
Nov1003, 07:23 PM

P: 640

to find horizontal *** you need to take the lim of the function as x approaches infinity.




#10
Nov1003, 10:03 PM

P: 1,572

or if you're in precalc, as the title of the thread implies, and a precalc class that doesn't use limits yet, then you could start with the statement given without proof that y=0 is a HA for the graph of G(t)=a^t, 0<a, 1!=a, and argue that your G shifts that HA in a certain way. by the way, this is what i'd do even if i knew about limits.



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