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Okay I have no idea how to solve this. Heres the problem: 

1. Plot an accurate graph of function f if the domain is -5 < x < 5. Function is f (x) = 1.7^x
2. Calculate f (3.721) Show that this point is on the graph.  
3. From your graph, approximately what is the value of x if f (x) = 4?  
4. Evaluate f (x) for the value of x you found in part 3. This value should be close to, but not quite equal to 4.  
5. Get a five significant digit approximation for the value of x in part (c). You may try short cuts in the iteration process.  
6. Find a five significant digit approximation for the value of x that makes f (x) = 0.5 

if someone could draw a graph with this as well, that would be awesome... thanks for your help! 
 
 Jan 18, 2015

Best Answer 

 #1
avatar+130536 
+5

Here's the graph........https://www.desmos.com/calculator/brwx1waam2

 

f(x) = 4  when x = about 2.6...proof 1.7^2.6 =  about 3.973

We could go through an iteration process, but here's a better way (hey, they said we could take shortcuts, didn't they??)

x = log(4) / log(1.7) = 2.612552871704276....I'm not that familiar with significant figures, so I'll let you trim as many off this as you need

And f(x) = 0.5 when x = about -1.3.....proof 1.7^(-1.3) = about 0.5016

Again....instead of an iteration process, use this

x = log(0.5) / log(1.7) = -1.306276435852138...again I'll let you worry about the sig figs

I think that's most of it....

 

 Jan 18, 2015
 #1
avatar+130536 
+5
Best Answer

Here's the graph........https://www.desmos.com/calculator/brwx1waam2

 

f(x) = 4  when x = about 2.6...proof 1.7^2.6 =  about 3.973

We could go through an iteration process, but here's a better way (hey, they said we could take shortcuts, didn't they??)

x = log(4) / log(1.7) = 2.612552871704276....I'm not that familiar with significant figures, so I'll let you trim as many off this as you need

And f(x) = 0.5 when x = about -1.3.....proof 1.7^(-1.3) = about 0.5016

Again....instead of an iteration process, use this

x = log(0.5) / log(1.7) = -1.306276435852138...again I'll let you worry about the sig figs

I think that's most of it....

 

CPhill Jan 18, 2015

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