Consider the following function. f(x) = x^2/5, a = 1, n = 3,...

Consider the following function. \(f(x) = x^{2/5}\), \(a = 1\), \(n = 3\), \(0.8 \le x \le 1.2\) (a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number \(a\). \(T_3(x) = \frac{8x^3}{125} - \frac{39x^2}{125} + \frac{104x}{125} + \frac{52}{125}\) (b) Use Taylor's Inequality to estimate the accuracy of the approximation \(f(x) \approx T_n(x)\) when \(x\) lies in the given interval. (Round your answer to eight decimal places.) \(|R_3(x)| \le 0.10000803\)

Transcript

00:01 Hello everyone, in this problem we are given with the function f of x to be equal to x power 3 by 5 with the value of a to be equal to 1 and n is equal to 3 and x lies in the interval 0 .8 less than or equal to x and what point 2.

00:19 So, now we need to approximate the given function by the taylor polynomial with the degree n at the number a.

00:27 So, here the function f of x is given as x power 3 by 5 and f dash of x is 3 divided by 5 into x power 2 by 5 and f double dash of x will be equal to minus 6 divided by 25 x power 7 by 5 and f triple dash of x to be equal to 42 divided by 125 x power 12 by 5.

01:15 So, thus so now, the taylor's polynomial of degree 3 of x will be equal to f of a plus f dash of a divided by 1 factorial into x minus a plus f double dash of a divided by 2 factorial into x minus a the whole square plus f triple dash of a divided by 3 factorial into x minus a the whole cube.

01:53 Now, substituting the value of a to be 1.

01:56 So, we get this taylor polynomial as f of 1 plus f dash of 1 divided by 1 factorial multiplied by x minus 1 plus f double dash of 1 divided by 2 factorial into x minus 1 the whole square plus f triple dash of 1 divided by 3 factorial into x minus 1 the whole cube.

02:24 So, now simplifying this we have this value to be f of 1 substituting the value of x to be 1 in the given f f of x oh it is 1 plus now substituting the value of x to be equal to 1 in f dash of x we get this value to be 3 by 5 multiplied by x minus 1 plus so substituting the value of x to be 1 in f double dash of x so simplifying the 2 factorial now we get the value to be minus 3 by 25 of x minus 1 the whole square plus now substituting the value of x to be equal to 1 in triple dash of x so we get this value to be 7 divided by 125 of x minus 1 the whole cube.

03:17 So, this is the required answer for the first part of the question.

03:24 Now, let us move on to the second part of the question...

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