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What is the answer for question 3. b) http://postimg.org/image/a55kae0gl/

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lkballer24 | Student, Grade 11 | Valedictorian

Posted July 31, 2013 at 10:41 PM via web

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What is the answer for question 3. b)

http://postimg.org/image/a55kae0gl/

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aruv | High School Teacher | Valedictorian

Posted August 1, 2013 at 4:28 AM (Answer #1)

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`1. (x_1,y_1)=((3pi)/8,0)`  and `(x_2,y_2)=((3pi)/8+pi/8,1)`

Slope of the secant passng through these two points=`(Y_2-y_1)/(x_2-x_1)`

`=1/((3pi)/8)=8/(3pi)`

`2. (x_1,y_1)=((3pi)/8,0) `  and `(x_2,y_2)=((16pi)/40,.27)`

Slope of the secant passng through these two points=`(y_2-y_1)/(x_2-x_1)=.27/(pi/40)`

`=10.8/pi`

3. ` (x_1,y_1)=((3pi)/8,0) ` and `(x_2,y_2)=((76pi)/200,.97)`

slope of the secant passng through these points is=`(.99)/(pi/200)``=198/pi`

4.`(x_1,y_1)=((3pi)/8,0)`    and `(x_2,y_2)=((376pi)/1000,.99)`

slope of the secant passng through these points is=`.99/(pi/1000)`

`=990/pi`

`f(x)-tan(2x)+1`

`f'(x)=2sec^2(2x)`

`f'(x)}_{x=3pi/8}=2sec^2(2xx(3pi)/8)=`

`=2xx2=4`

Thus equation of the tangent is

`y-0=4(x-(3pi)/8)`

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