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

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Unformatted text preview: 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 = m ( x- x ). For this case we have that: • y = f (2) = 2 4- 1 = 15 • x = 2 • m = f ( x ) = 4 x 3 = 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- . 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...
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This note was uploaded on 05/19/2011 for the course MATH 102 Math 102 taught by Professor Allard during the Winter '09 term at UBC.

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

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