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

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Homework Help Overview

The discussion revolves around the calculation of the derivative g'(2.5) using given values of g'(2) and g'(3). Participants are examining the validity of a local approximation method and the implications of using secant and tangent slopes in this context.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between secant and tangent slopes, questioning the assumption that the derivative at an intermediate point can be derived from the endpoints. There is discussion about the validity of linear interpolation as an approximation method.

Discussion Status

The conversation is ongoing, with participants providing insights into the limitations of the original poster's approach. Some participants suggest reconsidering the assumptions made about the function's behavior between the given points.

Contextual Notes

There is a focus on the distinction between the values of the function and its derivatives, as well as the implications of using linear interpolation in this scenario. The original poster's misunderstanding of the problem setup is acknowledged, but no resolution has been reached.

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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