Answer
$(f+g)(2k)=16k^2-6k-8$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To find the expression, $
(f+g)(2k)
,$ given that
\begin{array}{l}\require{cancel}
f(x)=3x^2-4 \text{ and }
\\g(x)=x^2-3x-4
,\end{array}
use the definition of the appropriate function operation. Then substitute $x$ with $2k.$
$\bf{\text{Solution Details:}}$
Using $(f+g)(x)=f(x)+g(x),$ then
\begin{array}{l}\require{cancel}
(f+g)(x)=(3x^2-4)+(x^2-3x-4)
\\\\
(f+g)(x)=3x^2-4+x^2-3x-4
\\\\
(f+g)(x)=(3x^2+x^2)-3x+(-4-4)
\\\\
(f+g)(x)=4x^2-3x-8
.\end{array}
Sustituting $x$ with $2k,$ then
\begin{array}{l}\require{cancel}
(f+g)(2k)=4(2k)^2-3(2k)-8
\\\\
(f+g)(2k)=4(4k^2)-3(2k)-8
\\\\
(f+g)(2k)=16k^2-6k-8
.\end{array}