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3 years ago

Solve:

Mathematics

2 answers:

harina [27]3 years ago

8 0

Answer:

$x = 0$

Step-by-step explanation:

Write the number in exponential form with the base of 2:

${2}^{x - 3} = {2}^{ - 3}$

Since the bases are the same, set the exponents equals.

$x - 3 = - 3$

Cancel equal terms on both sides of the equation:

$x = 0$

dedylja [7]3 years ago

5 0

<h2>Given</h2>

$2 {}^{x - 3} = \frac{1}{8}$

To Find value of 'x'

<h3><u>Solution</u></h3>

<u> $2 {}^{x - 3} = \frac{1}{8} \\ \\ \implies2 {}^{x - 3} = 2 {}^{ - 3} \\ \\ \\ Since \: Bases \: Are \\ Equal \: We \: Need \\ To \: Evaluate \\ The \: Powers \\ \\ \implies \: x - 3 = - 3 \\ \\ \implies x = - 3 + 3 \\ \\ \implies \: x = 0$ </u>

<u> $\fbox \red{Hope This Helps You ❤}$ </u>

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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

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Answer:

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

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[-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

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

In the particular case f

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

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

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

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

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

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

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

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]

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