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posted by  dancing missa on 11/2/2009 4:10:08 PM  |  status: Closed  |  Earned Karma: 25

limits, please help will rate

Course Textbook Chapter Problem Needs by
N/A Calculus Early Vectors Preliminary Editin by Stewart 3.1 2 11/2/2009 at 5:00:00 PM
Question Details:
if g(x)=1-x^3, find g'(0) and use it to find the equation of the tangent line to the curve y=1-x^3 at the point (0,1)
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posted by glengainer on 11/2/2009 4:32:48 PM  |  status: Live
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Response Details:
g(x)=1-x^3
First find g'(x)
g'(x)=3x^2 Using (g(x-h)-g(x))/h or points rule
Plug in the x term of the point (in this case 0)
g'(x)=3x^2
g'(x)=3(0^2)
g'(x)=0
So 0 is the slope of the tangent line to the curve y=1-x^3 
Then using plug in the slope and the point values into (m=slope)
y=mx+k
1=0*0+k
1=k
plug k back in
y=mx+k
So the answer is
y=1
Hope my answer is a lifesaver :) Feel free to PM me if you need more help
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posted by eswarramesh on 11/3/2009 12:12:06 PM  |  status: Live
Asker's Rating: Lifesaver   
Response Details:
given          g(x)=1-x^3
              ∴ g ' (x) = - 3 x^2         [  d/dx ( x^n) = n x^n-1 ]
          substitute  x=0 we get g(0) = - 3 (0) ^ 2 = 0
              given curve y= 1-x^3
             the derivative of the curve at the given point (0,1) gives the slope of the line
                hence (dy/dx) at (0,1) = -3 (0)^2 = 0
                   slope point form is y-y1=m (x-x1)     where (x1,y1) is the given point and m = slope the equation
                      hence y-1 = 0 (x-x1)=0
                         ∴ hence y=1 is the equation of the tangent
eswarramesh
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