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Bingel [31]
1 year ago
14

In which interval is the radical function f of x is equal to the square root of the quantity x squared plus 2 times x minus 15 e

nd quantity increasing?
[3, ∞)
(4, ∞)
[–5, 3]
(–∞, –5] ∪ [3, ∞)
Mathematics
1 answer:
sergij07 [2.7K]1 year ago
3 0

Using function concepts, it is found that it is increasing on the interval:

(–∞, –5] ∪ [3, ∞)

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

The function is given by:

f(x) = \sqrt{x^2 + 2x - 15}

The graph is given at the end of this question.

  • If the function is pointing upwards, it is increasing. Otherwise, it is decreasing.
  • In the graph, it can be seen that it is pointing upwards for x of -5 and less, or 3 and higher, thus, the interval is:

(–∞, –5] ∪ [3, ∞)

A similar problem is given at brainly.com/question/13539822

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If a(x) = 3x + 1 and b (x) = StartRoot x minus 4 EndRoot, what is the domain of (b circle a) (x)?
kiruha [24]

Answer:

[1,\infty)

Step-by-step explanation:

b(x)=\sqrt{x-4}

a(x)=3x+1

Since we want to know the domain of (b \circ a)(x), let's first consider the domain of the inside function, that is, that of a(x)=3x+1. Every polynomial function has domain all real numbers.

So we can plug anything for function a and get a number back.

Now the other function is going to be worrisome because it has a square root. You cannot take square root of negative numbers if you are only considering real numbers which that is the case with most texts.

Let's find (b \circ a)(x) and simplify now.

(b \circ a)(x)

b(a(x))

b(3x+1)

\sqrt{(3x+1)-4}

\sqrt{3x+1-4}

\sqrt{3x-3}

Now again we can only square root positive or zero numbers so we want 3x-3 \ge 0.

Let's solve this to find the domain of (b \circ a)(x).

3x-3 \ge 0

Add 3 on both sides:

3x \ge 3

Divide both sides by 3:

x \ge 1

So we want x to be a number greater than or equal to 1.

The option that says this is [1,\infty)

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

Give an example why option A fails:

A number in the given set is -2.

a(x)=3x+1

b(x)=\sqrt{x-4}

So a(-2)=3(-2)+1=-6+1=-5 and b(-5)=\sqrt{-5-4}=\sqrt{-9} \text{ which is not real}.

Give an example why option B fails:

A number in the given set is 0.

a(x)=3x+1

b(x)=\sqrt{x-4}

So a(0)=3(0)+1=0+1=1 and b(1)=\sqrt{1-4}=\sqrt{-3} \text{ which is not real}.

Give an example why option D fails:

While all the numbers in set D work, there are more numbers outside that range of numbers that also work.

A number not in the given set that works is 3.

a(x)=3x+1

b(x)=\sqrt{x-4}

So a(3)=3(3)+1=9+1=10 and b(1)=\sqrt{10-4}=\sqrt{6} \text{ which is real}.

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