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3 years ago

Using interval notation, the domain off is [- 3, 5] and the domain of g is [- 2, 7] . On what domain is f defined

Mathematics

1 answer:

Scorpion4ik [409]3 years ago

3 0

Note: I am assuming you mean ''on what domain is f+g is defined''.

Answer:

We conclude that

Dom(f+g) = [-2, 5]

Step-by-step explanation:

Given the domain of f

[-3, 5]

Given the domain of g

[-2, 7]

The domain of a composition f+g would include the intersection of both domains.

In other words, the domain of a composition f+g includes all the inputs of x-values that are in the domain of both functions i.e. f and g.

Domain of f = [-3, 5]

The x-values lie in the domain of f interval are:

x = -3, -2, 1, 0, 1, 2, 3, 4, 5

Domain of g = [-2, 7]

The x-values lie in the domain of g interval are:

x = -2, 1, 0, 1, 2, 3, 4, 5, 6, 7

Therefore, the common x-values in both f and g would be:

-2, 1, 0, 1, 2, 3, 4, 5

Thus, we conclude that

Dom(f+g) = [-2, 5]

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2 years ago

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3 years ago

Read 2 more answers

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Zina [86]

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A.10

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3 years ago

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2 years ago

Completing the square method sucks!!

slamgirl [31]

$\huge\underline\purple{Answer ☘}$

$4x {}^{2} + 3x + 5 = 0 \\ \\ dividing \: by \: 4... \\ \frac{4x {}^{2} }{4} + \frac{3x}{4} + \frac{5}{4} = 0 \\ x {}^{2} + \frac{3x}{4} + \frac{5}{4} = 0 \\ \\ we \: know \: that... \\ \: (a + b) {}^{2} = a {}^{2} + b {}^{2} + 2ab \\ \\ here... \\ 2ab = \frac{3x}{4} \\ = > 2xb = \frac{3x}{4} \: \: \: \: \: \: \: \: \: (a = x) \\ 2b = \frac{3}{4} \\ b = \frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$

<h3>nσw ín σur єquαtíσn... </h3>

$x {}^{2} + \frac{3x}{4} + \frac{5}{4} = 0 \\ \\adding \: and \: subtracting \: (\frac{3}{8}){}^{2} \\ \\ x {}^{2} + \frac{3x}{4} + \frac{5}{4} + ( \frac{3}{8} ) {}^{2} { - ( \frac{3}{8} })^{2} = 0 \\ \\ x {}^{2} + \frac{3x}{4} + (\frac{3}{8} ) {}^{2} + \frac{5}{4} - ( \frac{3}{8} ) {}^{2} = 0 \\ \\ (x + \frac{3}{8} ) {}^{2} + \frac{5}{4 } - ( \frac{3}{8} ) {}^{2} = 0 \\ \\ (x + \frac{3}{8} ) {}^{2} = (\frac{3}{8} ) {}^{2} - \frac{5}{4} \\ \\ (x + \frac{3}{8} ) {}^{2} = \frac{9}{64} - \frac{5}{4} \\ \\ (x + \frac{3}{8} ) {}^{2} = \frac{9 - 5(16)}{64} \\ \\ (x + \frac{3}{8} ) {}^{2} = \frac{9 - 80}{64} \\ \\ (x + \frac{3}{8} ) {}^{2} = \frac{ - 71}{64}$

<h3>ѕíncє.. </h3>

ѕquαrє σf αnч numвєr cαn't вє nєgαtívє..

ѕσ.. αnѕwєr dσєѕ nσt єхíѕt~

hσpє hєlpful~

<h2>~Be Brainly!</h2>

3 0

3 years ago