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(D) The equation $f(x)^{g(x)}=1$ is satisfied in the following cases:Case $1$: $f(x)=1$$x^2-7x+11=1 \implies x^2-7x+10=0 \implies (x-2)(x-5)=0 \implie

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$16$

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(D) The equation $f(x)^{g(x)}=1$ is satisfied in the following cases:Case $1$: $f(x)=1$$x^2-7x+11=1 \implies x^2-7x+10=0 \implies (x-2)(x-5)=0 \implies x=2, 5$.Case $2$: $g(x)=0$ and $f(x) 
eq 0$$x^2-6x-7=0 \implies (x-7)(x+1)=0 \implies x=7, -1$.Case $3$: $f(x)=-1$ and $g(x)$ is an even integer$x^2-7x+11=-1 \implies x^2-7x+12=0 \implies (x-3)(x-4)=0 \implies x=3, 4$.Check $g(x)$ for $x=3, 4$:For $x=3$,$g(3)=3^2-6(3)-7=9-18-7=-16$ (even,so $x=3$ is a solution).For $x=4$,$g(4)=4^2-6(4)-7=16-24-7=-15$ (odd,so $x=4$ is not a solution).The set of real values of $x$ is $\{2, 5, 7, -1, 3\}$.The sum of these values is $2+5+7-1+3=16$.

The sum of all the real values of $x$ satisfying the equation $(x^2-7x+11)^{x^2-6x-7}=1$ is

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