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Dovator [93]
3 years ago
12

If the function f (x) has a domain of (a,b] and a range of [c,d), then what is the domain and range of g (x) = m × f (x) + n?

Mathematics
1 answer:
xz_007 [3.2K]3 years ago
6 0
<h3>Answer: Choice A</h3>

Domain = (a,b]

Range = [mc + n,md + n)

==============================================

Explanation:

The domain stays the same because we still have to go through f(x) as our first hurdle in order to get g(x).

Think of it like having 2 doors. The first door is f(x) and the second is g(x). The fact g(x) is dependent on f(x) means that whatever input restrictions are on f, also apply on g as well. So going back to the "2 doors" example, we could have a problem like trying to move a piece of furniture through them and we'd have to be concerned about the f(x) door.

-------------------

The range will be different however. The smallest value in the range of f(x) is y = c as it is the left endpoint. So the smallest f(x) can be is c. This means the smallest g(x) can be is...

g(x) = m*f(x) + n

g(x) = m*c + n

All we're doing is replacing f with c.

So that means mc+n is the starting point of the range for g(x).

The ending point of the range is md+n for similar reasons. Instead of 'c', we're dealing with 'd' this time. The curved parenthesis says we don't actually include this value in the range. A square bracket means include that value.

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Step-by-step explanation:

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8 0
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Step-by-step explanation:

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4 0
3 years ago
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sineoko [7]

Answer:

\frac{1}{2\sqrt{5} }

Step-by-step explanation:

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\frac{d}{dx}\text{sin(y)}=\frac{1}{6}

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\frac{dy}{dx}=\frac{1}{6\text{cos(y)}}

cos(y) = \sqrt{1-\text{sin}^{2}(y) }

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Therefore, from equation (1),

\frac{dy}{dx}=\frac{1}{6\sqrt{1-\frac{x^2}{36}}}

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At x = 4,

\frac{d}{dx}[\text{sin}^{-1}(\frac{4}{6})]=\frac{1}{6\sqrt{1-\frac{4^2}{36}}}

\frac{d}{dx}[\text{sin}^{-1}(\frac{2}{3})]=\frac{1}{6\sqrt{1-\frac{16}{36}}}

                   =\frac{1}{6\sqrt{\frac{36-16}{36}}}

                   =\frac{1}{6\sqrt{\frac{20}{36} }}

                   =\frac{1}{\sqrt{20}}

                   =\frac{1}{2\sqrt{5}}

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4 years ago
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zheka24 [161]
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