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Linear approximations

  1. Mar 7, 2010 #1
    A non-linear device has the output input relation y = f(x) = x2 (x squared). Assuming the operating point is x0 = 1.5, the linear approximation for small changes would be given by:
    (a) y = 1.5x
    (b) y = x2 (x squared)
    (c) y = 3x
    (d) y = 3.5x
    (e) none of the above

    Can anyone tell me the answer? I was thinking it could be y = x squared?
  2. jcsd
  3. Mar 7, 2010 #2


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    A linear approximation L(x) would approximate the function f(x) with a line. In other words, around the point x=x0, you'd have [itex]L(x) = mx+b \approx f(x)[/itex]. Geometrically, L(x) is the line tangent to f(x) at x=x0. You're supposed to find the appropriate values of m and b to make that work.

    The answer (b) is the one you should immediately see is wrong since it's not a linear function.
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