Q. 32

Question

In Exercises 21–30 in Section 8.2 you were asked to find the fourth Maclaurin polynomial $P_{4} (x)$ for the specified function. In Exercises 23–32 we ask you to give Lagrange’s form for the corresponding remainder, $R_{4} (x)$ .

$x^{2} e^{x}$

Step-by-Step Solution

Verified

Answer

The remainder is $\frac{e^{c} (c^{2} + 8 c + 14)}{120} x^{5}$

1Step 1 : Given Information

Given equation : $x^{2} e^{x}$

Theory used : For $n > 0, if | f^{(n + 1)} (c) | \leq 1$ for every value of $x$ , then using the Lagrange's form for the remainder, we have

$R_{n} (x) = \frac{f^{(n + 1)} (c)}{(n + 1)!} x^{n + 1}$

So, the Lagrange's form for the remainder, $R_{4} (x) is$

$R_{4} (x) = \frac{f^{(5)} (c)}{5!} x^{5}$

2Step 2 : Calculating Lagrange’s form for the corresponding remainder

1) $f^{(1)} (x^{2} e^{x}) = x^{2} e^{x} + 2 x e^{x}$

2) $f^{(2)} (x^{2} e^{x}) = x^{2} e^{x} + 2 x e^{x} + 2 e^{x}$

3) $f^{(3)} (x^{2} e^{x}) = x^{2} e^{x} + 4 x e^{x} + 4 e^{x}$

4) $f^{(4)} (x^{2} e^{x}) = x^{2} e^{x} + 6 x e^{x} + 8 e^{x}$

5) $f^{(5)} (x^{2} e^{x}) = x^{2} e^{x} + 8 x e^{x} + 14 e^{x}$

So,

$R_{4} (x) = \frac{c^{2} e^{c} + 8 c e^{c} + 14 e^{c}}{5!} x^{5} = \frac{e^{c} (c^{2} + 8 c + 14)}{120} x^{5}$

Q. 31

Q.33

Other exercises in this chapter

Q. 30

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Q. 31

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Q.33

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Q. 34

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