The given functions are:
[tex]\begin{gathered} f(x)=2x^2-x+1 \\ g(x)=4x+3 \end{gathered}[/tex]Replace x with g(x) in the expression for the function f(x):
[tex]f(g(x))=2(g(x))^2-g(x)+1[/tex]Substitute the equation for g(x) into the equation above:
[tex]\begin{gathered} f(g(x))=2(4x+3)^2-(4x+3)+1_{} \\ \Rightarrow f(g(x))=2(16x^2+24x+9)-4x-3+1 \\ \Rightarrow f(g(x))=32x^2+48x+18-4x-3+1 \\ \Rightarrow f(g(x))=32x^2+48x-4x+18-3+1 \\ \Rightarrow f(g(x))=32x^2+44x+16 \end{gathered}[/tex]Notice that the composite function f(g(x)) is a quadratic equation.
To find the range of a quadratic equation, you have to find the maximum or the minimum value of the function.
The maximum or minimum point of a quadratic function is the y-coordinate of the vertex.
Note that if the coefficient of x² is positive, then the function has a minimum, but if the coefficient is negative, then the function has a maximum.
In this case, the function has a positive coefficient of x², hence, it has a minimum.
The x-coordinate of the vertex of a quadratic function f(x)=ax²+bx+c is given as:
[tex]x=-\frac{b}{2a}[/tex]Substitute b=44 and a=32 into the equation:
[tex]x=-\frac{44}{2(32)}=-\frac{11}{16}[/tex]Substitute x=-11/16 into the composite function to get the y-coordinate of the vertex:
[tex]y=32(-\frac{11}{16})^2+44(-\frac{11}{16})+16=\frac{7}{8}[/tex]Hence, since the function has a minimum as stated, the range will be all real numbers that are greater or equal to 7/8:
[tex]Range\colon\lbrack\frac{7}{8},\infty)[/tex]Find the function g(f(x)) using the same procedure:
[tex]\begin{gathered} g(x)=4x+3 \\ \text{Replace x with f(x):} \\ g(f(x))=4(f(x))+3 \\ \Rightarrow g(f(x))=4(2x^2-x+1)+3 \\ \Rightarrow g(f(x))=8x^2-4x+4+3 \\ \Rightarrow g(f(x))=8x^2-4x+7 \end{gathered}[/tex]Since
Since
