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Elan Coil [88]
3 years ago
14

Determine whether the function f(x) = 3(x − 1)4 is even or odd. The function is even because f(x) = f(−x). The function is odd b

ecause f(x) = f(−x). The function is even because f(x) = −f(x). The function is odd because f(x) = −f(x).
Mathematics
1 answer:
Ksivusya [100]3 years ago
4 0

Answer:

Step-by-step explanation:

given that

f(x)=3(x-1)^4

In order to determine whether a function is odd or even we are required to substitute x with -x and simplify

if f(-x) =f(x) it is an even function

if f(-x)=-f(x) it is an odd function

Hence

Let us put x=-x in our f(x)

f(-x)=3(-x-1)^4\\=3(-1(x+4))^4\\=3(x+1)^4

which is an entirely a  new function has no relation with the f(x) Hence it is neither an even function or an odd function.

I think some part of the question is missing here. Please recheck it once.

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A circle is translated 4 units to the right and then reflected over the x-axis. Complete the statement so that it will always be
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Answer:

The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

Step-by-step explanation:

Let C = (h,k) the coordinates of the center of the circle, which must be transformed into C'=(h', k') by operations of translation and reflection. From Analytical Geometry we understand that circles are represented by the following equation:

(x-h)^{2}+(y-k)^{2} = r^{2}

Where r is the radius of the circle, which remains unchanged in every operation.

Now we proceed to describe the series of operations:

1) <em>Center of the circle is translated 4 units to the right</em> (+x direction):

C''(x,y) = C(x, y) + U(x,y) (Eq. 1)

Where U(x,y) is the translation vector, dimensionless.

If we know that C(x, y) = (h,k) and U(x,y) = (4, 0), then:

C''(x,y) = (h,k)+(4,0)

C''(x,y) =(h+4,k)

2) <em>Reflection over the x-axis</em>:

C'(x,y) = O(x,y) - [C''(x,y)-O(x,y)] (Eq. 2)

Where O(x,y) is the reflection point, dimensionless.

If we know that O(x,y) = (h+4,0) and C''(x,y) =(h+4,k), the new point is:

C'(x,y) = (h+4,0)-[(h+4,k)-(h+4,0)]

C'(x,y) = (h+4, 0)-(0,k)

C'(x,y) = (h+4, -k)

And thus, h' = h+4 and k' = -k. The statement is now presented as:

\exists\, (h,k)\in \mathbb{R}^{2} /f: (x-h^{2})+(y-k)^{2}=r^{2}\implies f': [x-(h+4)]^{2}+[y-(-k)]^{2} = r^{2}

In other words, this mathematical statement can be translated as:

<em>There is a point (h, k) in the set of real ordered pairs so that a circumference centered at (h,k) and with a radius r implies a equivalent circumference centered at (h+4,-k) and with a radius r. </em>

<em />

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3 years ago
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