Asked by Greg
Using a graphing calculator and the concept of differentials determine the values for x when 1/(1+3x)^5 is approx = to (1-15x) within .01
Answers
Answered by
Steve
since the line 1-15x is the tangent line at (0,1), we could just find where
|1/(1+3x)^5) - (1-15x)| < 0.01
-0.008353 < x < 0.008872
Using differentials, we have
y(0) = 1
dy = -15/(1+3x)^6 dx
So, if we want |dy| < 0.01, we need x such that
1/(1+3x)^5 - (1-15x) < 0.01
which is just what we had above.
|1/(1+3x)^5) - (1-15x)| < 0.01
-0.008353 < x < 0.008872
Using differentials, we have
y(0) = 1
dy = -15/(1+3x)^6 dx
So, if we want |dy| < 0.01, we need x such that
1/(1+3x)^5 - (1-15x) < 0.01
which is just what we had above.
Answered by
Elodea
I need help please, how can I get a graphing calculator?