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2 years ago

Find the approximate solution of this system of equations.

1 answer:

Montano1993 [528]2 years ago

4 0

Y = |x² - 3x + 1|
y = x - 1

|x² - 3x + 1| = x - 1
|x² - 3x + 1| = ±1(x - 1)
|x² - 3x + 1| = 1(x - 1)   or    |x² - 3x + 1| = -1(x - 1)
|x² - 3x + 1| = 1(x) - 1(1) or |x² - 3x + 1| = -1(x) + 1(1)
|x² - 3x + 1| = x - 1     or         |x² - 3x + 1| = -x + 1
  x² - 3x + 1 = x - 1    or          x² - 3x + 1 = -x + 1
- x - x + x   + x
x² - 4x + 1 = -1       or          x² - 2x + 1 = 1
+ 1 + 1 - 1 - 1
x² - 4x + 1 = 0     or        x² - 2x + 0 = 0
x = -(-4) ± √((-4)² - 4(1)(1)) or x = -(-2) ± √((-2)² - 4(1)(0))
2(1)   2(1)
x = 4 ± √(16 - 4)        or        x = 2 ± √(4 - 0)
2 2
x = 4 ± √(12)    or      x = 2 ± √(4)
2 2
x = 4 ± 2√(3)       or         x = 2 ± 2
2 2
x = 2 ± √(3)     or          x = 1 ± 1
x = 2 + √(3) or x = 2 - √(3) or    x = 1 + 1 or x = 1 - 1
x = 2     or       x = 0
y = x - 1      or           y = x - 1 or y = x - 1 or y = x - 1
y = (2 + √(3)) - 1 or y = (2 - √(3)) - 1       or       y = 2 - 1 or y = 0 - 1
y = 2 - 1 + √(3)   or     y = 2 - 1 - √(3)      or           y = 1    or       y = -1
y = 1 + √(3) or      y = 1 - √(3) (x, y) = (2, 1) or (x, y) = (0, -1)
(x, y) = (2 ± √(3), 1 ± √(3))

The solution (0, -1) can be made by one function (y = x - 1) while the solution (2 ± √(3), 1 ± √(3)) can be made by another function (y = |x² - 3x + 1|). So the solution (2, 1) can be made by both functions, making the two solutions equal.

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Vladimir79 [104]

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Step-by-step explanation:

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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