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If $f(x)=ax^4-bx^2+x+5$ and $f(-3)=2$, then what is the value of $f(3)$?

Guest Oct 9, 2017
 #1
avatar+7155 
+2

Here's one way....

 

f(x)  =  ax4 - bx2 + x + 5                        Plug in  -x  for  x  .

 

f(-x)  =  a(-x)4 - b(-x)2 + (-x) + 5             (-x)4  =  x4     and     (-x)2  =  x2

 

f(-x)  =  ax4 - bx2 - x + 5                        Rearrange.

 

f(-x)  =  ax4 - bx2 + 5 - x                        Add  x  and subtract  x  .

 

f(-x)  =  ax4 - bx2 + x + 5 - x - x            Substitute  f(x)  in for  ax4 - bx2 + x + 5  .

 

f(-x)  =  f(x) - x - x          Plug in  3  for  x .

 

f(-3)  =  f(3) - 3 - 3         Substitute  2  in for  f(-3)  and solve for  f(3) .

 

2  =  f(3) - 3 - 3

 

2  =  f(3) - 6

 

8  =  f(3)

hectictar  Oct 9, 2017
 #2
avatar+87569 
+2

ax^4-bx^2+x+5$ and $f(-3)=2

 

a(-3)^4 - b(-3)^2 + -3 + 5  = 2

 

81a - 9b +  2  = 2

 

81a - 9b  = 0

 

9a  =  b

 

So  f(3)

 

a(3)^4 - 9a(3)^2+ 3 + 5 =

 

81a - 81a + 3 + 5

 

3 + 5   =  

 

8

 

 

cool cool cool

CPhill  Oct 9, 2017

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