Given:-
[tex]g(x)=x^2+3,h(x)=4x-3[/tex]To find:-
[tex](g-h)(1)[/tex]At first we find the value of (g-h)(x), so we get,
[tex]\begin{gathered} (g-h)(x)=g(x)-h(x) \\ =x^2+3-(4x-3) \\ =x^2+3-4x+3 \\ =x^2-4x+6 \end{gathered}[/tex]So the value of,
[tex](g-h)(x)=x^2-4x+6[/tex]So the value of (g-h)(1) is,
[tex]\begin{gathered} (g-h)(x)=x^2-4x+6 \\ (g-h)(1)=1^2-4\times1+6 \\ (g-h)(1)=1-4+6 \\ (g-h)(1)=7-4 \\ (g-h)(1)=3 \end{gathered}[/tex]So the required value is,
[tex](g-h)(1)=3[/tex]