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PLease

Veteran  Oct 21, 2017

Best Answer 

 #1
avatar+7336 
+3

(a)     Here's a graph:  https://www.desmos.com/calculator/fxto5gtfns

 

(b)     P  =  (2, 8)          Q  =  (x, x3)

 

Find the slope of the line  PQ  for the following values of  x .

 

When  x  =  1.5,   Q  =  (1.5, 1.53)  =  (1.5, 3.375)

 

We want to find the slope of the line passing through the points  (2, 8)  and  (1.5, 3.375)

 

slope  =  \(\frac{8-f(1.5)}{2-1.5}\,=\,\frac{8-3.375}{2-1.5}\,=\,\frac{4.625}{0.5}\,=\,9.25\)

 

This is the first value for the slope of the secant on the table.

 

x msec
   
1.5 9.25
   
1.9 \(\frac{8-f(1.9)}{2-1.9}\,=\,\frac{8-6.859}{2-1.9}\,=\,\frac{1.141}{0.1}\,=\,11.41\)
   
1.99         \(\frac{8-f(1.99)}{2-1.99}\,=\,\frac{8-7.880599}{2-1.99}\,=\,\frac{0.119401}{0.01}\,=\,11.9401\)
   
2 ?
   
2.01 \(\frac{8-f(2.01)}{2-2.01}\,=\,\frac{8-8.120601}{2-2.01}\,=\,\frac{-0.120601}{-0.01}\,=\,12.0601\)
   
2.1 \(\frac{8-f(2.1)}{2-2.1}\,=\,\frac{8-9.261}{2-2.1}\,=\,\)

 

Do you see how I am doing these? See if you can finish the table...

 

(c)

 

The slope of the tangent line when  x = 2  appears to be between  11.9401  and  12.0601

 

Let's just take the average of these two...                  (11.9401  +  12.0601) / 2  =  12.0001

 

So I'd guess that the slope of the tangent line at  (2, 8)  is  12    laugh

hectictar  Oct 21, 2017
edited by hectictar  Oct 21, 2017
edited by hectictar  Oct 21, 2017
 #1
avatar+7336 
+3
Best Answer

(a)     Here's a graph:  https://www.desmos.com/calculator/fxto5gtfns

 

(b)     P  =  (2, 8)          Q  =  (x, x3)

 

Find the slope of the line  PQ  for the following values of  x .

 

When  x  =  1.5,   Q  =  (1.5, 1.53)  =  (1.5, 3.375)

 

We want to find the slope of the line passing through the points  (2, 8)  and  (1.5, 3.375)

 

slope  =  \(\frac{8-f(1.5)}{2-1.5}\,=\,\frac{8-3.375}{2-1.5}\,=\,\frac{4.625}{0.5}\,=\,9.25\)

 

This is the first value for the slope of the secant on the table.

 

x msec
   
1.5 9.25
   
1.9 \(\frac{8-f(1.9)}{2-1.9}\,=\,\frac{8-6.859}{2-1.9}\,=\,\frac{1.141}{0.1}\,=\,11.41\)
   
1.99         \(\frac{8-f(1.99)}{2-1.99}\,=\,\frac{8-7.880599}{2-1.99}\,=\,\frac{0.119401}{0.01}\,=\,11.9401\)
   
2 ?
   
2.01 \(\frac{8-f(2.01)}{2-2.01}\,=\,\frac{8-8.120601}{2-2.01}\,=\,\frac{-0.120601}{-0.01}\,=\,12.0601\)
   
2.1 \(\frac{8-f(2.1)}{2-2.1}\,=\,\frac{8-9.261}{2-2.1}\,=\,\)

 

Do you see how I am doing these? See if you can finish the table...

 

(c)

 

The slope of the tangent line when  x = 2  appears to be between  11.9401  and  12.0601

 

Let's just take the average of these two...                  (11.9401  +  12.0601) / 2  =  12.0001

 

So I'd guess that the slope of the tangent line at  (2, 8)  is  12    laugh

hectictar  Oct 21, 2017
edited by hectictar  Oct 21, 2017
edited by hectictar  Oct 21, 2017
 #2
avatar+280 
+3

See i was doing this exact thing, but why does that formula have the (4), where does it come from? Other than thanks for the break down, i was trying to figure this out, and if it wasnt for that for it would have been fine. cause no matter what i put in it it, i was getting answer of 1.

Veteran  Oct 21, 2017
 #3
avatar+7336 
+2

The formula at the top confused me too.

 

I don't know for sure what it is supposed to mean, but if it said

 

\(m_{sec}=\frac{f(x)-f(2)}{x-2}\)                  then I think it would make sense.

hectictar  Oct 21, 2017
edited by hectictar  Oct 21, 2017

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