The functions are given as,
[tex]\begin{gathered} f(x)=3x+2 \\ g(x)=x^2+x-6 \\ h(x)=-2x\cdot f(x)-g(x) \end{gathered}[/tex]
Substitute the values of f(x) and g(x) in h(x), and simplify,
[tex]\begin{gathered} h(x)=-2x\cdot(3x+2)-(x^2+x-6) \\ h(x)=-2x\cdot(3x)-2x\cdot(2)-x^2-x+6 \\ h(x)=-6x^2-4x-x^2-x+6 \\ h(x)=-7x^2-5x+6 \end{gathered}[/tex]
Thus, option C is the correct choice as it represents the correct expression of the function h(x).
Consider the given function g(x) as,
[tex]g(x)=-\sqrt[]{x+3}+2[/tex]
Since the function f(x) is not given here, we will use the same function from the above part of the problem,
[tex]f(x)=3x+2[/tex]
According to the given condition,
[tex]f(x)=g(x)[/tex]
Substitute the values,
[tex]\begin{gathered} 3x+2=-\sqrt[]{x+3}+2 \\ 3x=-\sqrt[]{x+3} \end{gathered}[/tex]
Consider this as the parent equation.
Square both sides,
[tex]\begin{gathered} 9x^2=x+3 \\ 9x^2-x-3=0 \end{gathered}[/tex]
Apply the quadratic formula,
[tex]\begin{gathered} x=\frac{-(-1)\pm\sqrt[]{(-1)^2-4(9)(-3)}}{2(9)} \\ x=\frac{1\pm\sqrt[]{1+108}}{18} \\ x=\frac{1+\sqrt[]{109}}{18},\frac{1-\sqrt[]{109}}{18} \end{gathered}[/tex]
Consider that the square root term cannot contain any negative number inside. So the following solution is not feasible,
[tex]x=\frac{1-\sqrt[]{109}}{18}[/tex]
So, the only feasible solution of the parent equation is,
[tex]x=\frac{1+\sqrt[]{109}}{18}[/tex]
Thus, there is only one value of 'x' that satisfies the condition.
Therefore, option B is the correct choice.
