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gogolik [260]
3 years ago
11

Find the domain x^2-25/2

Mathematics
1 answer:
Mumz [18]3 years ago
8 0
For this, I am assuming that -25/2 is not a part of the exponent(although in future questions you should make this clear by using parenthesis.

When I do problems like this, I come to CBAD.  where the variables are y=A(Bx+C)+D

CBAD helps significantly in domain and range problems where there is a well known parent function.  (y=x^2, y=sqrt(x), y=1/x, y=x, etc).  Basically, if there is a parent function with slight additions(y=2*x^2, y=sqrt(x+1), y=1/(5x), y=3x, etc)

This works that in the order, a C term shifts the graph horizontally in the opposite direction expected (a negative number will shift right, a positive number will shift left).  For example, if C is -2, the graph of the parent function will shift right 2.
Then, a B term dilates(changes the horizontal size of the graph) by the factor 1/B(every x value is multiplied by 1/B).  Thus a B >1 will shrink the graph horizontally, and a 0<B<1 with stretch the graph horizontally.  If B is negative, it reflects the graph over the y axis and then either shrinks or stretches horizontally.  If B is 1, nothing changes.
Then, a A term dilates(changes the horizontal size of the graph) by the factor A(every y value is multiplied by A).  Thus a A >1 will stretch the graph vertically, and a 0<A<1 with shrink the graph vertically.  If A is negative, it reflects the graph over the x axis and then either shrinks or stretches vertically.  If A is 1, nothing changes.
Finally, a D term shifts the graph vertically in the direction expected (a negative number will shift down, a positive number will shift up).  For example, if D is -2, the graph of the parent function will shift down 2.

<u>Warning </u><u></u><u>CBAD must be performed in the order C-B-A-D</u>

Now that you know these simple rules and can memorize the few parent functions, you can figure out the domain and range of almost any function.  

In your case, the function x^(2)-25/2 is not much different than x^2, an upwards facing parabola centered at (0,0).  In fact, there is only one transformation, D, because C=0 (it is not shown), B=1(it is not shown), and A=1(it is not shown).  Thus the only difference between your graph and the parent function is that yours is shifted down by 25/2.  

For this reason, because the domain of x^2 is not restricted (it includes all real numbers), the domain of your new function is not restricted (it includes all real numbers).
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0

2

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1 = 1/2(0) + 1

2 = 1/2(2) + 1

3 = 1/2(4) + 1

4 = 1/2(6) + 1

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djverab [1.8K]

Answer:

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Step-by-step explanation:

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8/21

Step-by-step explanation:

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Find a particular solution to the nonhomogeneous differential equation y′′+4y=cos(2x)+sin(2x).
I am Lyosha [343]
Take the homogeneous part and find the roots to the characteristic equation:

y''+4y=0\implies r^2+4=0\implies r=\pm2i

This means the characteristic solution is y_c=C_1\cos2x+C_2\sin2x.

Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form y_p=ax\cos2x+bx\sin2x. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.

With y_1=\cos2x and y_2=\sin2x, you're looking for a particular solution of the form y_p=u_1y_1+u_2y_2. The functions u_i satisfy

u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx
u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx

where W(y_1,y_2) is the Wronskian determinant of the two characteristic solutions.

W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2

So you have

u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx
u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x

u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx
u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x

So you end up with a solution

u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x

but since \cos2x is already accounted for in the characteristic solution, the particular solution is then

y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x

so that the general solution is

y=C_1\cos2x+C_2\sin2x-\dfrac14x\cos2x+\dfrac14x\sin2x
7 0
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Which expression is equal to x−5x2+3−x2−x+4x2+3 ?
insens350 [35]

the third option. I think

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