+0

# Idk this one

+1
71
2

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
Sort:

#1
+5565
+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
+79881
+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

CPhill  Oct 9, 2017

### 7 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details