Wednesday

July 30, 2014

July 30, 2014

Posted by **Hebe** on Tuesday, September 26, 2006 at 8:08pm.

I don't really get what this question is asking.

It looks like the area of right triangle to me...try the graph, and shade the area under the tangent.

Find out where the tangent line crosses the x-axis. Then draw yourself a figure to see what area they are talking about. The tangent line will intersect the x axis at some x value that is >0. I would suggest doing this by calculating the area under the y = x^2 curve and the x axis from x=0 to x=1, using a simple integral from 0 to 1 of y(x), and subtracting the triangular area formed by the tangent line, x=1 and the x axis.

Please ignore the "#1 of 2" in my name. There is aome problem with the Jiskha software automatically entering my name wrong when I prepare answers

i need a refresher on how to find the tangent line. thanks.

The first derivative of a function evaluated at the point will give the slope of the line tangent to the function there. Since you know the point and the slope you can determine the equation for the line.

Briefly

m=f'(x_o) at (x_o,f(x_o)) and the equation for the tangent line is

y-f(x_o)=f'(x_o)(x-x_o)

where (x_o,f(x_o)) is any point where f(x) is defined.

ok. I found the equation to be y=2x-1. I draw the diagram but i am not sure which area i am suppose to find. The triangle under the tangent line and the x-axis??

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