Local Approximation Mistake: g'(2.5)=-3

AI Thread Summary
The discussion centers on a misunderstanding of how to approximate the derivative g'(2.5) given the values g'(2) = 1 and g'(3) = -2. The mistake lies in assuming that the derivative at an intermediate point can be calculated using the difference in derivatives at the endpoints without considering the specific point of interest. It is emphasized that the approximation should involve linear interpolation rather than a direct calculation of the slope between the two endpoints. The correct approach for estimating g'(2.5) is to recognize that there are multiple possible values for derivatives at that point, not just a linear assumption based on endpoints. The key takeaway is that without additional information about the function's behavior, one cannot accurately determine g'(2.5) solely from g'(2) and g'(3).
UrbanXrisis
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when:
g'(2)=1
g'(3)= -2

msec=mtan
g'(2.5)=(y2-y1)/(x2-x1)
=(-2-1)/(3-2)
=-3/1
=-3

I got this question wrong on a test, were was my mistake?
 
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The primes on the function indicate the tangents at a point.

So f(x) is the value of the function at the point x = x and f'(x) is the value of the tangent at the point x = x. You were asked to find the tangent at an intermediate point given the tangents at the two extremeties of a subinterval [2,3]. You claim (in your solution) that the derivative at any intermediate point equals the ratio of the difference in derivatives at the end points of the subinterval to the length of the subinterval. Now think about this: your expression does not involve the point x = 2.5 anywhere. So it would be just as well if I could replace 2.5 by 2.75 and use your expression...which would mean that the tangents (or slopes) of the curve at both x = 2.5 and x = 2.75 are equal! Is that so? Not unless the function is constant in the subinterval [2.5, 2.75] but that again is not something you can assume.

Also what is msec = mtan?
 
the slope of secant = the slope of the tangent
 
Is that what you call a local approximation? Hmm...why do you think its wrong then?
 
The fact that these are derivatives is irrelevant. We are given that a function, f (which happens to be g' in this problem) at x= 2 and x= 3 and asked to "approximate" f(2.5). Given g'(2) and g'(3) there is NO way of saying for sure what g'(2.5) is but there are an infinite number of different "approximations" which may or may not be accurate.

The SIMPLEST approximation, given two points, is "linear interpolation". Since f(2)= 1 and f(3)= -2, the linear interpolation is just the number halfway between those:
(1+ (-2))/2= -1/2.

UrbanXrisis: IF you were given g(2)= 1 and g(3)= -2 (Not g' ) THEN the simplest estimate for g' anywhere between 2 and 3 would be the slope of the straight line:
(-2-1)/(2-1)= -3 but that was NOT what you were given!
 
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