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3 years ago

14

PLEASE HELP ME!!!!!!!

1 answer:

Zepler [3.9K]3 years ago

3 0

Honestly I don’t know it sorry

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Find the third order maclaurin polynomial. Use it to estimate the value of sqrt1.3

vodka [1.7K]

$\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$ is the maclaurin polynomial and estimate value of $\sqrt{1.3}$ is 1.14. This can be obtained by using the formula to find the maclaurin polynomial.

<h3>Find the third order maclaurin polynomial:</h3>

Given the polynomial,

$f(x)=\sqrt{1+3x}=(1+3x)^{\frac{1}{2} }$

The formula to find the maclaurin polynomial,

$f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}$

Next we have to find f'(x), f''(x) and f'''(x),

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

$f''(x) =-\frac{9}{4}(1+3x)^{-\frac{3}{2} }$

$f'''(x) = \frac{81}{8}(1+3x)^{-\frac{5}{2} }$

By putting x = 0 , we get,

$f(0)=(1+3(0))^{\frac{1}{2} }=1$

$f'(0) = \frac{3}{2}(1+3(0))^{-\frac{1}{2} }=\frac{3}{2}$

$f''(0) =-\frac{9}{4}(1+3(0))^{-\frac{3}{2} }=-\frac{9}{4}$

$f'''(0) = \frac{81}{8}(1+3(0))^{-\frac{5}{2} }=\frac{81}{8}$

Therefore the maclaurin polynomial by using the formula will be,

$\sqrt{1+3x}=f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^{2} + \frac{f'''(0)}{3!}x^{3}$

$\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$

To find the value of $\sqrt{1.3}$ we can use the maclaurin polynomial,

$\sqrt{1.3}$ is $\sqrt{1+3x}$ with x = 1/10,

$\sqrt{1+3(1/10)}=1+\frac{3}{2} (1/10)-\frac{9}{8} (1/10)^{2} + \frac{81}{8}(1/10)^{3}$

$\sqrt{1+3(1/10)}=\frac{18247}{16000} = 1.14$

Hence $\sqrt{1+3x}=1+\frac{3}{2} x-\frac{9}{8} x^{2} + \frac{81}{8}x^{3}$ is the maclaurin polynomial and estimate value of $\sqrt{1.3}$ is 1.14.

Learn more about maclaurin polynomial here:

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What is the value of b in the equation (y)*

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Apply the exponent rule that $(x^{m})^{n} = x^{mn}$ so now you get:

$y^{4b} = y^{-24}$

Since the base is y for both sides of the equation, just set the exponents equal to each other and solve for b.

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