Can someone help please?

Find T5(x) : Taylor polynomial of degree 5 of the function f(x)=cos(x) at a=0 . (You need to enter function.) T5(x)= Find all va

Burka [1]

Answer:

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

The polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

Step-by-step explanation:

According to Taylor's theorem

$\bf f(x)=f(0)+f'(0)x+f''(0)\displaystyle\frac{x^2}{2}+f^{(3)}(0)\displaystyle\frac{x^3}{3!}+f^{(4)}(0)\displaystyle\frac{x^4}{4!}+f^{(5)}(0)\displaystyle\frac{x^5}{5!}+R_6(x)$

with

$\bf R_6(x)=f^{(6)}(c)\displaystyle\frac{x^6}{6!}$

for some c in the interval (-x, x)

In the particular case f

f(x)=cos(x)

we have

$\bf f'(x)=-sin(x)\\f''(x)=-cos(x)\\f^{(3)}(x)=sin(x)\\f^{(4)}(x)=cos(x)\\f^{(5)}(x)=-sin(x)\\f^{(6)}(x)=-cos(x)$

therefore

$\bf f'(x)=-sin(0)=0\\f''(0)=-cos(0)=-1\\f^{(3)}(0)=sin(0)=0\\f^{(4)}(0)=cos(0)=1\\f^{(5)}(0)=-sin(0)=0$

and the polynomial approximation of T5(x) of cos(x) would be

$\bf cos(x)\approx1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{4!}=\\\\=1-\displaystyle\frac{x^2}{2}+\displaystyle\frac{x^4}{24}$

In order to find all the values of x for which this approximation is within 0.002652 of the right answer, we notice that

$\bf R_6(x)=-cos(c)\displaystyle\frac{x^6}{6!}$

for some c in (-x,x). So

$\bf |R_6(x)|\leq|\displaystyle\frac{x^6}{6!}|=\displaystyle\frac{|x|^6}{6!}$

and we must find the values of x for which

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652$

Working this inequality out, we find

$\bf \displaystyle\frac{|x|^6}{6!}\leq0.002652\Rightarrow |x|^6\leq1.90944\Rightarrow\\\\\Rightarrow |x|\leq\sqrt[6]{1.90944}\Rightarrow |x|\leq1.113826815$

Therefore the polynomial is an approximation with an error less than or equals to 0.002652 for x in the interval

[-1.113826815, 1.113826815]

8 0

2 years ago

10 POINTS AND WILL GIVE BRAINLIEST

Vladimir [108]

${3}^{ - 2x} = { (\frac{1}{ {3}^{2} } )}^{x} = ( { \frac{1}{9} )}^{x} an \\ exponential \: function \: of \: the \: form \\ f(x) = a \times {b}^{x} \: with \: a \: = 1 \: and \\ b = \frac{1}{9}$

8 0

3 years ago

Abit is planning a birthday party for his grandfather. He bought a cake for \$18$18. He also wants to buy some balloons, which a

Temka [501]

Answer:

3$

Step-by-step explanation: if he try to buy as many balloons as passible with the money he has it would be 3$

6 0

2 years ago

Zero of the functions f(x)= x^2 + 4x+3 and why ?

slava [35]

There are two zeros: -3 and -1. This is because when you factor f(x)= x^2 + 4x + 3, it becomes f(x) = (x + 3)(x + 1). In order to find the zeros, set f(x) to 0, and then solve for x for both (x + 3) and (x + 1). You can check your answer by substituting either -3 or -1 for x in the equation 0 = x^2 + 4x + 3.

6 0

2 years ago

Jessie pays $2 for each carnival ride plus $8

ikadub [295]

$ 2 each carnival ride + $8 spend at most $35

Rewrite that is
$2x + 8 ≤ 35

I hope this helped and that I was right, have a nice day.

3 0

3 years ago