Linear approximations

  • Thread starter mf42
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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?
 

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  • #2
vela
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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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