Chapter 9.7, Problem 12E

### Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095

Chapter
Section

### Calculus

10th Edition
Ron Larson + 1 other
ISBN: 9781285057095
Textbook Problem

# Conjecture Consider the function f ( x ) = x 2 e x .(a) Find the Maclaurin polynomials P 2 , P 3 , and P 4 for f.(b) Use a graphing utility to graph f , P 2 , P 3 and P 4 (c) Evaluate and compare the values of f ( n ) ( 0 ) and P n ( n ) ( 0 ) for n = 2 , 3 , and 4.(d) Use the results in part (c) to make a conjecture about f ( n ) ( 0 ) and P n ( n ) ( 0 )

(a)

To determine

To calculate: The Maclaurin polynomials P2,P3,P4 for the function given in the question, that is, f(x)=x2ex.

Explanation

Given:

The given function is f(x)=x2ex.

Formula used:

The definition of Taylor series for a polynomial of the nth degree is as follows:

If f has n number of derivatives at c, then the polynomial is called the nth Taylor series for f at c. In this case, if c=0 then the polynomial is called the Maclaurin series. It has been shown below.

Pn(x)=f(0)+f(c)(x)+f(0)2(x)2++f(n)(0)n(x)n

The derivative of the function, ex is given below.

ddx(ex)=ex

The derivative of the function, xn is given below.

ddx(xn)=nxn1

Calculation:

The function is,

f(x)=x2ex

Fid the derivatives of the function given above, with respect to x.

f(x)=ddx(x2ex)

Apply the product rule, that is ddx(uv)=vdudx+udvdx, in order to evaluate the aforementioned derivative.

f(x)=exddx(x2)+x2ddx(ex)

Make use of the formula, ddx(ex)=ex and ddx(xn)=nxn1 in order to obtain the derivative as follows:

f(x)=ex(2x)+x2(ex)=ex(x2+2x)

Put x=0 in the above derivative, and simplify as shown below,

f(x)=e0((0)2+2(0))=1×0=0

Again, differentiate, f(x)=ex(x2+2x) with respect to x.

f(x)=ddx(ex(x2+2x))

Apply the product rule, that is ddx(uv)=vdudx+udvdx in order to evaluate the above derivative.

f(x)=(x2+2x)ddx(ex)+exddx(x2+2x)

Make use of the formula, ddx(ex)=ex and ddx(xn)=nxn1 in order to obtain the above derivative,

f(x)=(x2+2x)(ex)+ex(2x+2)=ex(x2+4x+2)

Put x=0 in the f(x)

f(0)=e0((0)2+4(0)+2)=1(0+0+2)=2

Differentiate f(x)=ex(x2+4x+2) with respect to x.

f(x)=ddx(ex(x2+4x+2))

Use the product rule, that is ddx(uv)=vdudx+udvdx in order to evaluate the above derivative as follows:

f(x)=(x2+4x+2)ddx(ex)+exddx(x2+4x+2)

Make use of the formula, ddx(ex)=ex and ddx(xn)=nxn1 in order to evaluate the above derivative

f(x)=(x2+4x+2)(ex)+ex(2x+4)=ex(x2+6x+6)

Put x=0 in the above derivative,

f(4)(x)=e0((0)2+6(0)+6)=1(6)=6

Differentiate f(x)=ddx(ex(x2+6x+6)) with respect to x.

f(4)(x)=ddx(ex(x2+6x+6))

Use the product rule, ddx(uv)=vdudx+udvdx in order to evaluate the above derivative

(b)

To determine

To graph: The Maclaurin polynomials, which are P2,P3,P4, and the function, f(x)=x2ex.

(c)

To determine

To calculate: The value of fnn(0) and Pnn(0) in case of f(x)=x2ex when n=2,3,4

(d)

To determine
The conjecture about, fnn(0) and Pn(n)(0), from the result obtained in part (c) when n=2,3,4 as:

 n fnn(0) Pn(n)(0) 2 2 2 3 6 6 4 12 12

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