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pantera1 [17]
3 years ago
8

2. Confirm that 1+V5 and 1 - V5 are the correct roots for x2 – 2x – 4 = 0.

Mathematics
1 answer:
hoa [83]3 years ago
5 0

Answer:

see explanation

Step-by-step explanation:

Given

x² - 2x - 4 = 0 ( add 4 to both sides )

x² - 2x = 4

To solve use the method of completing the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 1)x + 1 = 4 + 1

(x - 1)² = 5 ( take the square root of both sides )

x - 1 = ± \sqrt{5} ( add 1 to both sides )

x = 1 ± \sqrt{5}, thus

x = 1 + \sqrt{5} or x = 1 - \sqrt{5}

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Why do the functions f(x) = sin−1(x) and g(x) = cos−1(x) have different ranges?
Gnom [1K]
We Know that
For a function to have an inverse function, it must be one-to-one—that is, it must pass the Horizontal Line Test.

1. On the interval [–pi/2, pi/2], the function y = sin x is increasing
2. On the interval [–pi/2, pi/2], y = sin x takes on its full range of values, [–1, 1]
3. On the interval [–pi/2, pi/2], y = sin x is one-to-one
sin x has an inverse function on this interval [–pi/2, pi/2]

On the restricted domain [–pi/2, pi/2]  y = sin x has a unique inverse function called the inverse sine function. <span>f(x) = sin−1(x)
</span>the range of y=sin x  in the domain [–pi/2, pi/2]  is [-1,1] 
the range of y=sin-1  x  in the domain [-1,1]  is [–pi/2, pi/2]  

1. On the interval [0, pi], the function y = cos x is decreasing
2. On the interval [0, pi], y = cos x takes on its full range of values, [–1, 1]
3. On the interval [0, pi], y = cos x is one-to-one
cos x has an inverse function on this interval [0, pi]

On the restricted domain [0, pi]  y = cos x has a unique inverse function called the inverse sine function. f(x) = cos−1(x)
the range of y=cos x  in the domain [0, pi]  is [-1,1] 
the range of y=cos-1  x  in the domain [-1,1]  is [0, pi] 

the answer is

<span>the values ​​of the range are different because the domain in which the inverse function exists are different</span>  
8 0
3 years ago
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