Question

hw 11 - chain rule section 2.6: problem 1 (1 point)
let
$f(x)=10x^{2}+4$ and $g(x)=4x^{2}-1$.
then
$f(f(x))=square$,
$f(g(x))=square$, and
$g(f(x))=square$.

Explanation:

Step1: Find $f(f(x))$

Substitute $f(x)$ into $f(x)$. Given $f(x)=10x^{2}+4$, then $f(f(x)) = 10(10x^{2}+4)^{2}+4$. Expand $(10x^{2}+4)^{2}=100x^{4}+80x^{2}+16$. So $f(f(x))=10(100x^{4}+80x^{2}+16)+4=1000x^{4}+800x^{2}+160 + 4=1000x^{4}+800x^{2}+164$.

Step2: Find $f(g(x))$

Substitute $g(x)$ into $f(x)$. Given $g(x)=4x^{2}-1$, then $f(g(x))=10(4x^{2}-1)^{2}+4$. Expand $(4x^{2}-1)^{2}=16x^{4}-8x^{2}+1$. So $f(g(x))=10(16x^{4}-8x^{2}+1)+4=160x^{4}-80x^{2}+10 + 4=160x^{4}-80x^{2}+14$.

Step3: Find $g(f(x))$

Substitute $f(x)$ into $g(x)$. Given $f(x)=10x^{2}+4$, then $g(f(x))=4(10x^{2}+4)^{2}-1$. Expand $(10x^{2}+4)^{2}=100x^{4}+80x^{2}+16$. So $g(f(x))=4(100x^{4}+80x^{2}+16)-1=400x^{4}+320x^{2}+64 - 1=400x^{4}+320x^{2}+63$.

Answer:

$f(f(x)) = 1000x^{4}+800x^{2}+164$
$f(g(x)) = 160x^{4}-80x^{2}+14$
$g(f(x)) = 400x^{4}+320x^{2}+63$