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Use the Quadratic Approximation Error Bound to bound the error |f(x)−T2(x)| on the interval [-1,1]. True or false: The Quadratic Approximation Error Bound indicates that |f(x)−T2(x)|≤ 0.004307 for all x in I. True

Answer :

tatendagota

Answer:

[tex]T(2)_{x} = 1 + \frac{1}{3} x + \frac{1}{18} x^{2}[/tex]

Step-by-step explanation:

Thinking process:

Let the function be: f(x)−T2(x)| on the interval [-1,1].

Then, if [tex]f(x) =e^{\frac{x}{3} }[/tex]

Then [tex]f(x)^{'1} = \frac{1}{3}e^{\frac{1}{3} }[/tex]

         [tex]f^{11}(x) = \frac{1}{9}e^{\frac{1}{3} } , then f^{11}/2 = 2![/tex]

       and so on..

Then the function becomes:

[tex]T(2) x = 1 + \frac{1}{3}x + \frac{1}{18}x^{2}[/tex]

okpalawalter8

The resultant function of the Quadratic Approximation Error Bound is mathematically given as

[tex]T(2) x = 1 + \frac{1}{3}x + \frac{1}{18}x^{2}[/tex] , The correct option is True(A).

Does the Quadratic Approximation Error Bound indicate that |f(x)−T2(x)|≤ 0.004307 for all x in I?

Question Parameters:

the Quadratic Approximation Error Bound to bound the error |f(x)−T2(x)| on the interval [-1,1]

Generally, the equation for the Function   is mathematically given as

f(x)−T2(x)

Therefore

for [tex]f(x) =e^{\frac{x}{3} }[/tex]

The function gives

[tex]f(x)^{'1} = \frac{1}{3}e^{\frac{1}{3} }[/tex]

[tex]f^{11}/2 = 2*1[/tex]

f^{11}/2 = 2

In conclusion, The resultant function becomes

[tex]T(2) x = 1 + \frac{1}{3}x + \frac{1}{18}x^{2}[/tex]

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