\"Just need the answers\" Approximation: \"Just need the answers\" Use linear ap
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Question
"Just need the answers"
Approximation:
"Just need the answers"
Use linear approximation, i.e. the tangent line, to approximate as follows: Let f(x) = x The equation of the tangent line to f(x) x = 81 can be written in the form y = mx + b where m is: and where b is: Using this, we find our approximation for 81.2 is Approximate 3 22.8 by means of differentials or a linear approximation. Use for the base point the closest integer that is convenient with regard to the cube root function. Warning: WeBWorK wants the approximation found using the method described above; a root read off a calculator display will be counted as wrong.Explanation / Answer
(a)
f(x) = sqrt(x)
f'(x) = 1/2 x^-1/2
f(x+h) = sqrt(81.2) = sqrt(81 + 0.2)
Here h = 0.2, x = 81
f(x) = 9
f'(x) = [f(x + h) - f(x)]/h
putting x = 9 in f'(x) we get
1/6 = [f(x+h) - 9]/0.2
f(x + h) - 9 = 0.2/6
f(x + h) = 9.033
Hence sqrt(81.2) = 9.033
(b)
f(x) = x^1/3
f'(x) = 1/3 x^-2/3
f(x + h) = (22.8)^1/3 = (27 - 4.2)^1/3
here x = 27, h = -4.2
Hence f(x) = 3 & f'(x) = 1/27
f'(x) = [f(x + h) - f(x)]/h
1/27 = [f(x + h) - 3]/(-4.2)
f(x + h) - 3 = -4.2/27
f(x + h) = 2.844
Hence cuberoot (22.8) = 2.844
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