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

In summary, the mistake in the question was assuming that the derivative at any intermediate point equals the ratio of the difference in derivatives at the endpoints of the subinterval to the length of the subinterval. This is incorrect as the function may not be constant in the subinterval. The correct way to approximate the derivative at an intermediate point is through linear interpolation, which takes the average of the values at the endpoints. The equation msec=mtan refers to the fact that the slope of the secant line between two points is equal to the slope of the tangent at one of those points. This is not a reliable method of approximation as it does not take into account the shape of the function.
  • #1
UrbanXrisis
1,196
1
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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  • #2
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?
 
  • #3
the slope of secant = the slope of the tangent
 
  • #4
Is that what you call a local approximation? Hmm...why do you think its wrong then?
 
  • #5
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!
 

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

What is a local approximation mistake?

A local approximation mistake refers to an error in the process of estimating a value using a local approximation method. This can occur when trying to approximate a value using a tangent line or a linear approximation, and the actual value differs from the estimated value.

What does g'(2.5)=-3 mean?

g'(2.5)=-3 is a notation used to represent the derivative of the function g at the point x=2.5. In this case, the value of the derivative at x=2.5 is -3.

How is g'(2.5)=-3 calculated?

To calculate g'(2.5)=-3, we use the definition of a derivative which states that the derivative at a point is equal to the slope of the tangent line at that point. In this case, we would need to find the slope of the tangent line at x=2.5 and if it is equal to -3, then g'(2.5)=-3.

Why is it important to be aware of local approximation mistakes?

Being aware of local approximation mistakes is important because it can help us avoid making incorrect conclusions or decisions based on inaccurate estimations. It is crucial for accurate mathematical modeling and problem solving.

How can we minimize local approximation mistakes?

One way to minimize local approximation mistakes is by using more precise and accurate methods of approximation, such as using higher order derivatives or numerical methods. It is also important to be aware of the limitations of local approximation and to check for accuracy and reasonableness of the estimated values.

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