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DIA [1.3K]
1 year ago
5

What is the range of a piecewise function g of x, where part 1 is x squared minus 5 for x less than negative 3 and part 2 is 2 t

imes x for x greater than or equal to negative 3.? (–[infinity], [infinity]) [–6, [infinity]) [–5, [infinity]) [5, [infinity])
Mathematics
1 answer:
svetoff [14.1K]1 year ago
3 0

The range of a piecewise function g(x) for x < -3 and x\geq-3 is ( -6, ∞ ).

     

Given function,

g(x)=\left \{ {{x^2-5 ~for~x ~ < ~-3 } \atop {2x~for~x~\geq-3}} \right.
We have to find the range of this function.

If we take x < -3 we have to take the function x^2-5.

If we take x~\ge-3 we have to take the function 2x.

<h3>What is the range of a function?</h3>

The set of the outputs of a function is called the range of a function.

Example: f(x) = 3x + 1 where x = 1,2,3

Range of f(x) = {4, 7, 10}.

We see that the x values we can take for the given function are:

x < -3\\x\geq-3

We see that x can take values as:

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

We can say also write as x ∈ ( -∞, ∞ ).

Let's take x values of x\geq-3

For x = -3, g(x) = 2 x -3 = -6.

For x = -2, g(x) = 2 x -2 = -4.

For x = -1, g(x) = 2 x -1 = -2.

For x = 0, g(x) = 2 x 0 = 0

For x = 1, g(x) = 2 x 1 = 2.

For x = 2, g(x) = 2 x 2 = 4 and so on.

If we take x \geq-3 we see that g(x) will tend to  + ∞.

Lets take x values of x < -3.

For x = -4, g(x) = -4 x -4  - 5 = 16 - 5 = 11.

For x = -5, g(x) = -5x-5 -  5  = 25 - 5 = 20.

For x = -6, g(x) = -6 x -6 -  5 = 36 - 5 = 31.

For x = -7 g(x) = -7 x -7 - 5 = 49 - 5 = 44 and so on.

If we take x < -3 we see that g(x) tend to + ∞.

Thus for the given x values, the lowest value of g(x) is -6, and g(x) values keep on increasing.

Thus the range of a piecewise function g(x) for x < -3 and x\geq-3 is ( -6, ∞ ).      

Learn more about the range of a function here:

brainly.com/question/21027387

#SPJ1

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Answer:17,000

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Read 2 more answers
If f(p) divided by x-p and x-q have the same remainder
oee [108]
Hello,


x^2-y^2=(x+y)(x-y)
x^3-y^3=(x-y)(x²+xy+y²)

Let's use Horner's division

.........|a^3|a^2.|a^1..........|a^0
.........|1....|5....|6..............|8....
a=p...|......|p....|5p+p^2....|6p+5p^2+p^3
----------------------------------------------------------
.........|1....|5+p|6+5p+p^2|8+6p+5p^2+p^3

The remainder is 8+6p+5p^2+p^3 or 8+6q+5q^2+q^3


Thus:
8+6p+5p^2+p^3 = 8+6q+5q^2+q^3
==>p^3-q^3+5p^2-5q^2+6p-6p=0
==>(p-q)(p²+pq+q²)+5(p-q)(p+q)+6(p-q)=0
==>(p-q)[p²+pq+q²+5p+5q+6]=0 or p≠q
==>p²+pq+q²+5p+5q+6=0

And here, Mehek are there sufficients explanations?
3 0
3 years ago
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