We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
47
4
avatar+1129 

1.What is the upper bound of the function?

f(x)=4x4−25x2−5x−13

Enter your answer in the box.

upper bound =

 

2.

What is the end behavior of the polynomial function?

Drag the choices into the underlines to correctly describe the end behavior of the function. The option for both could be either  f(x)→−∞ or f(x)→∞

                        f(x)=−7x3−x+1                                

As x→−∞                                                                 

As x→−∞    


                       f(x)=2x4+6x2+4

 

As x→∞

As x→∞
 

 Sep 10, 2019
 #1
avatar+103061 
+1

1.What is the upper bound of the function?

f(x)=4x^4−25x^2−5x−13

 

I'll admit....I had this in Pre-Cal  ages ago, but I forgot how to calculate the upper and lower bounds on the zeroes without a littlte "refresher"....here are the steps to finding the upper [ and lower] bounds on the zeroes:

 

1. The leading coefficient must be a "1".....if not.....divide  all the terms by the leading coefficient.....so we have

x^4   - (25/4)x^2 - (5/4) - (13/4)

 

2.  Write down all the coefficients   →   1, (-25/4) , (-5/4) , (-13/4)

 

3. "Throw away" the lead coefficient , 1

 

4. Remove the minus signs  ....so...we are left with the values   25/4, 5/4 , 13/4

 

5. We now have two  possible bounds :

 

(a)  Bound 1  =  the largest value + 1  =  (25/4) + 1  =  (25/4) + (4/4)  =   29/4

(b)  Bound 2  = the sum of all values or 1, whichever is larger.....so...

25/4  + 5/4  + 13/4  =  43/4    which is larger than 1

 

The smallest of these two possible bounds is our answer  =  29/4  =  7.25

 

This tells us that  that the upper  and lower values on zeroes  are  -7.25   and 7.25

 

So....the upper bound is  7.25

 

Look at the graph here:  https://www.desmos.com/calculator/wdwslnqzqa

 

The function has two real roots.....notice that they are contained between  x = -7.25  and x  =7.25

 

I do not know why this works....I expect that the proof of it is difficult  !!!!

 

BTW :  here is a good site that reviews this : https://www.mathsisfun.com/algebra/polynomials-bounds-zeros.html

 

 

cool cool cool

 
 Sep 10, 2019
edited by CPhill  Sep 10, 2019
 #3
avatar+1129 
0

wouldnt than answer be just 13 since whn you do long divsion  13/4 -25 -5 -13 it is 4 27 346 and 4485 which are all postitive ?

 
jjennylove  Sep 12, 2019
 #2
avatar+103061 
+1

Second one

 

f(x)  = -7x^3  - x + 1

Remember the rule .....if the lead coefficient is negative  and the power on its associated variable is odd:

The poynomial  will tend toward   + infinity   as x  →  - infinity

And....the polynomial will tend tword -infinity as x →   infinity

See the graph here : https://www.desmos.com/calculator/y47dnl4gy5

 

f(x)  =  2x^4 + 6x^2 + 4

The lead coefficient is positive and the power on its associated variable is even

The polynomial will approach + infinity   as x → -infinity  and as x →  infinity

See the graph here :  https://www.desmos.com/calculator/lmgxbyiyzo

 

 

cool cool cool

 
 Sep 10, 2019
 #4
avatar+1129 
+1

thanks for your help! I found out that the answer for the last one was actaully postitive infinty as well. Just ated to let you know so you can make  note of it.Thanks again though for your help i aprreacite it.

 

f(x)  =  2x^4 + 6x^2 + 4

The lead coefficient is positive and the power on its associated variable is even

The polynomial will approach + infinity   as x → infinity  and as x →  infinity

 
jjennylove  Sep 12, 2019

32 Online Users

avatar
avatar
avatar
avatar
avatar
avatar