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3 years ago

Find the domain for the function f(x)=sqrt x^2-x+6

Mathematics

1 answer:

dybincka [34]3 years ago

6 0

Answer:

The domain is {x : x ∈ R} or (-∞ , ∞)

Step-by-step explanation:

* Lets explain how to find the domain

- The domain of the function is the values of x which make the

function defined

- The quantity under the square root must be ≥ 0 because there is

no square root for negative value

* Lets solve the problem

∵ f(x) = √(x² - x + 6)

∴ The value of (x² - x + 6) must be greater than or equal zero because

there is no square root for negative value

- Graph the function to know which values of x make the quantity

under the root is negative that means the values of x which make

the graph under the x-axis

∵ The graph doesn't intersect the x-axis at any point

∵ All the graph is above the x-axis

∴ There is no value of x make f(x) < 0

∴ x can be any real number

∴ The domain of f(x) is all real numbers

∴ The domain is {x : x ∈ R} or (-∞ , ∞)

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shtirl [24]

<h3>Answer 1a :- </h3>

Here we can solve this question using<em> </em><em>Unitary</em><em> </em><em>Method</em><em> </em><em>.</em> As ;

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→ In 1hr the boat will travel 452km/17

→ In 9hrs the boat will travel 452/17km * 9 = 239.29km

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→ It will travel 1km on 13/156L of gasoline .

→ It will travel 480km on 13/156*480L = 40L of gasoline .

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<h3>Answer 2a :-</h3>

→ A boat can go 180 miles on 60L gasoline.

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<h3>Answer 2b :-</h3>

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2 years ago

How many nonzero terms of the Maclaurin series for ln(1 x) do you need to use to estimate ln(1.4) to within 0.001?

Vilka [71]

Answer:

The estimate of In(1.4) is the first five non-zero terms.

Step-by-step explanation:

From the given information:

We are to find the estimate of In(1 . 4) within 0.001 by applying the function of the Maclaurin series for f(x) = In (1 + x)

So, by the application of Maclurin Series which can be expressed as:

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2 f"(0)}{2!}+ \dfrac{x^3f'(0)}{3!}+... \ \ \ \ \ --- (1)$

Let examine f(x) = In(1+x), then find its derivatives;

f(x) = In(1+x)

$f'(x) = \dfrac{1}{1+x}$

f'(0) $= \dfrac{1}{1+0}=1$

f ' ' (x) $= \dfrac{1}{(1+x)^2}$

f ' ' (x) $= \dfrac{1}{(1+0)^2}=-1$

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

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

f ' ' ' '(x) $= \dfrac{6}{(1+x)^4}$

f ' ' ' '(x) $= \dfrac{6}{(1+0)^4}=-6$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+x)^5} = 24$

f ' ' ' ' ' (x) $= \dfrac{24}{(1+0)^5} = 24$

Now, the next process is to substitute the above values back into equation (1)

$f(x) = f(0) + \dfrac{xf'(0)}{1!}+ \dfrac{x^2f' \ '(0)}{2!}+\dfrac{x^3f \ '\ '\ '(0)}{3!}+\dfrac{x^4f '\ '\ ' \ ' \(0)}{4!}+\dfrac{x^5f' \ ' \ ' \ ' \ '0)}{5!}+ ...$

$In(1+x) = o + \dfrac{x(1)}{1!}+ \dfrac{x^2(-1)}{2!}+ \dfrac{x^3(2)}{3!}+ \dfrac{x^4(-6)}{4!}+ \dfrac{x^5(24)}{5!}+ ...$

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

To estimate the value of In(1.4), let's replace x with 0.4

$In (1+x) = x - \dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\dfrac{x^5}{5}- \dfrac{x^6}{6}+...$

$In (1+0.4) = 0.4 - \dfrac{0.4^2}{2}+\dfrac{0.4^3}{3}-\dfrac{0.4^4}{4}+\dfrac{0.4^5}{5}- \dfrac{0.4^6}{6}+...$

Therefore, from the above calculations, we will realize that the value of $\dfrac{0.4^5}{5}= 0.002048$ as well as $\dfrac{0.4^6}{6}= 0.00068267$ which are less than 0.001