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Dave uses the tangent line approximation on a certain function f(x) and obtains the approximations f(2.98) is about 2.08 and f(3.01) is about 1.96. What does the tangent line approximation give for f(2.99)?

 Aug 6, 2021
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I suppose you are learning interpolation .... a type of extimation that assumes the graph is linear between two points (2.98 and 3.01 in this case)

 

from (f(2.98) to f (3.01) is     1.96 - 2.08 =  - .12   

 

   (3.01 - 2.98)  is .03

 

-.12/.03 = -4      the graph changes   -4    per   1 change in f(x)

 

for   .01   change (from   2.98 to 2.99)

       it will be   2.08  - 4(.01) = 2.04  

 Aug 7, 2021

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