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What is the answer and how do you solve for question 5.b?...

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lkballer24 | Student, Grade 11 | Valedictorian

Posted April 22, 2013 at 7:06 PM via web

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What is the answer and how do you solve for question 5.b?

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rcmath | High School Teacher | (Level 1) Associate Educator

Posted April 22, 2013 at 8:17 PM (Answer #1)

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The function is `f(x)=0.25*4^x-2`
by looking at the answer for the first part of the question (please see the reference link below), we notice that as the values are getting closer to zero the slope is decreasing. 
So we will have to asuume that the slope will eventually decrease to zero. So your tangant will be a horizontal line.
That can be proved by looking at the limit as x approches zero of 
`(0.25*4^x-2-(0.25-2))/(x-0)=(.25*4^x-2-0.25+2)/x=(.25*4^x-0.25)/x` , the limit of such a function is 0.


 

 

 

 

Sources:

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oldnick | Valedictorian

Posted April 23, 2013 at 12:00 AM (Answer #2)

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`f(x)=1/4 4^x -2=4^(x-1)-2` `=2^(2(x-1))-2`

so: 

`f'(x)=2^(2x-1) log2` 

So  `m= f'(0)= 1/2log2`  and  `f(0)= -7/4`

tangent straigth line equation is

`y-y_0=m(x-x_0)`

`y-7/4=1/2 x log2 `

`y=1/4(2xlog2-7)`

Red line: function

Blue line: tangent straight line

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