Question

Suppose f(x)=ax^4+bx^2+x+5. Find f(4) given f(-4)=3

Answer 1

Let's just plug in \(-4:~f(-4)=4^4a+4^2b-4+5 = 3\)
Let \(u=4^4a+4^2b,\) we have \(f(-4)=u+1=3\)

So \(\underline{u = 2}\)

Now plug in \(4\), we have
\[\begin{array}~f(4) &= 4^4a+4^2b+4+5\\~\\ &= u+4+5\\~\\&=2+4+5\\~\\&=\boxed{11}\end{array}\]

Answer 2

What's u?

Answer 3

Just placehold for \(4^4a+4^2b\).

Answer 4

if you plug in -4, wouldn't it be \[-4a^4-4b^2 \] instead of \[4^4a+4^2b\]

Answer 5

Actually, it would be \((-4)^4a+(-4)^2b\) which is equal to \(4^4a+4^2b\) because even exponent cancels minus sign.

Answer 6

Do you understand what I did? Ask question if you are confused/

Answer 7

Yes, thank you so much!