Question

Find the degree 3 taylor polynomial t3(x) of function f(x)=(7x+2)3/2 at a=2. t3(x)=

Answer 1

Consider this option.

Answer 2

The degree 3 Taylor polynomial t3(x) of function f(x) is $(\frac{-343}{1024} ){(x-2)^{3} }$.      

What is a Taylor series?

Taylor series is a power series that gives the expansion of a function f (x) in the neighborhood of a point, provided that in the neighborhood the function is continuous, all its derivatives exist, and the series converges to the function.

For the given situation,

The function is f(x)=(7x+2)^3/2 at a=2.

The Taylor series expansion is

${\displaystyle f(a)+{\frac {f'(a)}{1!}}(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f'''(a)}{3!}}(x-a)^{3}+\cdots }$

Here t3(x) = ${\frac {f'''(a)}{3!}}(x-a)^{3}$

Now, ${f'(x)}=(7)(\frac{3}{2} )(7x+2)^{\frac{1}{2} }$

⇒ ${f'(x)}=(\frac{21}{2} )(7x+2)^{\frac{1}{2} }$

${f''(x)}=(7)(\frac{21}{2} )(\frac{1}{2}) (7x+2)^{\frac{-1}{2} }$

⇒ ${f''(x)}=(\frac{147}{4} ) (7x+2)^{\frac{-1}{2} }$

${f'''(x)}=(\frac{147}{4} ) (\frac{-1}{2} )(7x+2)^{\frac{-3}{2} }$

⇒ ${f'''(x)}=(\frac{-1029}{8} )(7x+2)^{\frac{-3}{2} }$

Then, ${f'''(2)}=(\frac{-1029}{8} )(7(2)+2)^{\frac{-3}{2} }$

⇒ ${f'''(2)}=(\frac{-1029}{8} )(16)^{\frac{-3}{2} }$

⇒ ${f'''(2)}=(\frac{-1029}{8} )(\frac{1}{64} )$

⇒ ${f'''(2)}=(\frac{-1029}{512} )$

Thus,  t3(x) = ${\frac {f'''(2)}{3!}}(x-2)^{3}$

⇒ $(\frac{-1029}{512} )\frac{(x-2)^{3} }{6}$

⇒ $(\frac{-343}{1024} ){(x-2)^{3} }$

Hence we can conclude that the degree 3 Taylor polynomial t3(x) of function f(x) is $(\frac{-343}{1024} ){(x-2)^{3} }$.

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