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Nezavi [6.7K]
3 years ago
5

Need help finding the end behavior !

Mathematics
1 answer:
Allushta [10]3 years ago
7 0

Let's focus on f(x) = |x| for now.

Recall that the absolute value of any number is never negative.

Some examples: |-7| = 7 and |5| = 5

So as x gets bigger in the positive direction, so does y. That explains the notation x \to \infty, \ f(x) \to \infty. Informally, we can say "the graph rises to the right".

Similarly, we have x \to -\infty, \ f(x) \to \infty which means it "rises to the left". Both endpoints rise to positive infinity. The left side of the graph goes up forever because again the result of any absolute value function is never negative. So if we plug in say negative a million, then the result is positive a million. In a sense, the V shape absolute value function is almost like a parabola. Both have the exact same end behavior on both sides.

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

Now let's move onto g(x) = 2x^2+4

The only thing that matters when determining the end behavior is the leading term. The leading term here is 2x^2

The even exponent means the endpoints either A) go up together or B) go down together. We go with case A because the leading coefficient is positive. Like I mentioned earlier, this parabola mimics the V shaped absolute value graph in terms of the end behaviors being the same.

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

Lastly, let's focus on h(y) = 3y^4-2

I'm not sure why your teacher is using y when the others were using x. I'll just swap y for x to get h(x) = 3x^4-2

Like the g(x) function, the largest exponent is even, so the left and right end behaviors go in the same direction. The positive leading coefficient means we have the endpoints going upward toward positive infinity.

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answer: she will have 8 sticks and $24 to replace

Step-by-step explanation:

4 drum kits x 2 drumsticks each kit = 8

8 drumsticks x $3 each = $24

<h2>HOPE IT HELP LET ME KNOW IF ANYTHING IS WRONG :D</h2>
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A

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The indicated function y1(x) is a solution of the associated homogeneous equation. Use the method of reduction of order to find
9966 [12]

Answer:

<em>The particular integral of given differential equation</em>

<em>                  </em>y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))<em></em>

<em> General solution of given differential equation</em>

<em>      </em>y = y_{c} + y_{p}<em></em>

<em>  </em>Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))<em></em>

<em></em>

Step-by-step explanation:

<u><em>Step(i)</em></u>:-

Given Differential equation  y'' − 5 y' + 4 y = x

Given equation in operator form

        D²y - 5 Dy +  4 y = x

⇒     ( D² - 5 D +  4 ) y =x

⇒    f(D) y = Q

where  f(D) = D² - 5 D +  4 and Q(x) = x

<em>The auxiliary equation  f(m) =0</em>

<em>           m²-5 m + 4 =0</em>

         m² - 4 m - m + 4 =0

        m ( m -4 ) -1 ( m-4) =0

         (m - 1) =0   and ( m-4) =0

        <em> m = 1 and m =4</em>

<em>The complementary function </em>

<em></em>Y_{c} = C_{1} e^{x} + C_{2} e^{4x}<em></em>

<u><em>Step(ii)</em></u>:-

<u><em>particular integral</em></u>

<em>Particular integral</em>

<em>     </em>y_{p} = \frac{1}{f(D)} Q(x) = \frac{1}{D^{2}  - 5 D +  4} X<em></em>

<em>taking common '4' </em>

<em>                          </em>= \frac{1}{4(1 +  (\frac{D^{2}  - 5 D}{4} ))} X<em></em>

<em>                         </em>

<em>                           </em>=\frac{1}{4}  (1 + (\frac{D^{2} -5D}{4})^{-1} )} X<em></em>

<em>applying binomial expression</em>

<em>      ( 1 + x )⁻¹    = 1 - x + x² - x³ +.....       </em>

<em>                          </em>=\frac{1}{4}  (1 - (\frac{D^{2} -5D}{4}) +((\frac{D^{2} -5D}{4})^{2} -...} )X<em></em>

<em>Now simplifying and we will use notation D = </em>\frac{dy}{dx}<em></em>

<em>                        </em>=\frac{1}{4}  (x - (\frac{D^{2} -5D}{4})x +((\frac{D^{2} -5D}{4})^{2}(x) -...}<em></em>

<em>Higher degree terms are neglected</em>

<em>                     </em>=\frac{1}{4}  (x - (\frac{ -5 D}{4}) x)<em></em>

<em>The particular integral of given differential equation</em>

<em>                  </em>y_{p} = \frac{1}{4} ( x - (\frac{-5}{4} ) (1))<em></em>

<u><em>Final answer</em></u><em>:-</em>

<em>          General solution of given differential equation</em>

<em>      </em>y = y_{c} + y_{p}<em></em>

<em>  </em>Y (x) = C_{1} e^{x} + C_{2} e^{4x} + \frac{1}{4} ( x + (\frac{5}{4} ))<em></em>

<em></em>

<em></em>

<em>         </em>

<em> </em>

     

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