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2 years ago

O. Find the domain and range of each of the three restricted functions.

Mathematics

1 answer:

Orlov [11]2 years ago

4 0

The domain and range of the functions sin x, cos x, and tan x will be (-∞, ∞) and [-1, 1], (-∞, ∞) and [-1, 1], and R - (2n + 1)π/2 and (-∞, ∞) respectively.

<h3>What are domain and range?</h3>

The domain means all the possible values of x and the range means all the possible values of y.

The three restricted functions are given below.

f(x) = sin (x)

f(x) = cos (x)

f(x) = tan (x)

Then the domain of the three restricted functions will be

The domain of sin x will be (-∞, ∞).

The domain of cos x will be (-∞, ∞).

The domain of tan x will be R - (2n + 1)π/2.

Then the range of the three restricted functions will be

The range of sin x will be [-1, 1].

The range of cos x will be [-1, 1].

The range of tan x will be (-∞, ∞).

The graph is given below.

More about the domain and range link is given below.

brainly.com/question/12208715

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3 years ago

Determine all prime numbers a, b and c for which the expression a ^ 2 + b ^ 2 + c ^ 2 - 1 is a perfect square .

kogti [31]

Answer:

The family of all prime numbers such that $a^{2} + b^{2} + c^{2} -1$ is a perfect square is represented by the following solution:

$a$ is an arbitrary prime number. (1)

$b = \sqrt{1 + 2\cdot a \cdot c}$ (2)

$c$ is another arbitrary prime number. (3)

Step-by-step explanation:

From Algebra we know that a second order polynomial is a perfect square if and only if $(x+y)^{2} = x^{2} + 2\cdot x\cdot y + y^{2}$ . From statement, we must fulfill the following identity:

$a^{2} + b^{2} + c^{2} - 1 = x^{2} + 2\cdot x\cdot y + y^{2}$

By Associative and Commutative properties, we can reorganize the expression as follows:

$a^{2} + (b^{2}-1) + c^{2} = x^{2} + 2\cdot x \cdot y + y^{2}$ (1)

Then, we have the following system of equations:

$x = a$ (2)

$(b^{2}-1) = 2\cdot x\cdot y$ (3)

$y = c$ (4)

By (2) and (4) in (3), we have the following expression:

$(b^{2} - 1) = 2\cdot a \cdot c$

$b^{2} = 1 + 2\cdot a \cdot c$

$b = \sqrt{1 + 2\cdot a\cdot c}$

From Number Theory, we remember that a number is prime if and only if is divisible both by 1 and by itself. Then, $a, b, c > 1$ . If $a$ , $b$ and $c$ are prime numbers, then $2\cdot a\cdot c$ must be an even composite number, which means that $a$ and $c$ can be either both odd numbers or a even number and a odd number. In the family of prime numbers, the only even number is 2.