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xz_007 [3.2K]
3 years ago
12

If , f(x)= 3/(x+2) minus the square root of x-3, complete the following statement: f(19) = _____ (need help as soon as possible

not sure if the answer is just 19?)
Mathematics
2 answers:
Svet_ta [14]3 years ago
4 0
All you need to do is plug 19 in for x and solve.

f(19) = 3/(19+2) - sqrt(19-3) = 1/7 - sqrt(16)= 1/7 - 4

The answer is -3.8571 or as a fraction -27/7

Natasha_Volkova [10]3 years ago
4 0

Answer:

f(x)= \frac{3}{x+2}- \sqrt{x-3}

the given function has domain

x+2≠0

x≠ -2

x-3≥0

x≥3

That is , x ∈ [3,∞]

Now, calculating ,f (19)

f(19)= \frac{3}{19+2}-\sqrt{19-3}\\\\ = \frac{3}{21} - \sqrt{16}\\\\ f(19)=\frac{1}{7}-4\\\\=\frac{1-28}{7}\\\\=\frac{-27}{7}

f(19)=-[3\frac{6}{7}]

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2 years ago
use Taylor's Theorem with integral remainder and the mean-value theorem for integrals to deduce Taylor's Theorem with lagrange r
Vadim26 [7]

Answer:

As consequence of the Taylor theorem with integral remainder we have that

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \int^a_x f^{(n+1)}(t)\frac{(x-t)^n}{n!}dt

If we ask that f has continuous (n+1)th derivative we can apply the mean value theorem for integrals. Then, there exists c between a and x such that

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}dt = \frac{f^{(n+1)}(c)}{n!} \int^a_x (x-t)^n d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{n+1}}{n+1}\Big|_a^x

Hence,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{n!} \frac{(x-t)^{(n+1)}}{n+1} = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1} .

Thus,

\int^a_x f^{(n+1)}(t)\frac{(x-t)^k}{n!}d t = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}

and the Taylor theorem with Lagrange remainder is

f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.

Step-by-step explanation:

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3 years ago
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Answer:

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Step-by-step explanation:

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