midterm1_solns

# midterm1_solns - Midterm 1 SOLUTIONS(50 points total Math...

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Midterm 1 SOLUTIONS (50 points total) Math 184, section 102 Date: September 28, 2007 Instructor: Ignacio Rozada 1. Use a linear approximation on the function f ( x ) = ( x 2 - 1)( x 2 + 1) to approximate f (2 . 01). Would you expect an approximation for f (1 . 99) to be more, or less precise than the one for f (2 . 01)? Notice first that f ( x ) = ( x 2 - 1)( x 2 + 1) = x 4 - 1. We want to find the equation of the tangent line to f ( x ) at x = 2, as it’s very close to the point that we want to approximate: x = 2 . 01. The equation of a line is y - y 0 = m ( x - x 0 ). For this case we have that: y 0 = f (2) = 2 4 - 1 = 15 x 0 = 2 m = f 0 ( x 0 ) = 4 x 3 0 = 32 Plugging everything in we get that y = 15 + 32( x - 2). Substituting in x = 2 . 01 yields that the linear approximation is 15 . 32, which is actually accurate to three digits. With respect to the accuracy that we should expect with an approximation to x = 1 . 99 ( y (1 . 99) = 15+32(1 . 99 - 2) = 15 - 0 . 32), the graph is growing faster at 2 . 01 than it is at 1 . 99, in other words at 1 . 99 we are closer to the point of slope zero, and because of this, the approximation at x = 1 . 99 will be slightly more precise than the one at 2 . 01, even though they are both the same distance from the point we picked to find the tangent line.

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