To evaluate both equations using a table of values, choose the values of x and solve each equation for these x-values.
Let's choose:
x = -2.25
x = -1.75
x = 0.50
x = 0.75
Given:
[tex]\begin{gathered} f(x)=\frac{x-1}{x^2+x-1} \\ \\ g(x)=3^x-2 \end{gathered}[/tex]
Let's substitute the values of x in each equation, solve them and compare the results using a table.
Let's begin with f(x).
[tex]\begin{gathered} f(x)=\frac{x-1}{x^{2}+x-1} \\ \\ f(-2.25)=\frac{-2.25-1}{(-2.25)^2-2.25-1}=\frac{-3.25}{5.06-3.25}=-1.79 \\ f(-1.75)=\frac{-1.75-1}{(-1.75)^2-1.75-1}=\frac{-2.75}{3.06-2.75}=-8.87 \\ f(0.5)=\frac{0.5-1}{(0.5)^2+0.5-1}=\frac{-0.5}{0.25-0.5}=2 \\ f(0.75)=\frac{0.75-1}{(0.75)^2+0.75-1}=\frac{-0.25}{0.56-0.25}=-0.8 \end{gathered}[/tex]
Now, let's evaluate g(x).
[tex]\begin{gathered} g(x)=3^{x}-2 \\ g(-2.25)=3^{-2.25}-2=-1.92 \\ g(-1.75)=3^{-1.75}-2=-1.85 \\ g(0.5)=3^{0.5}-2=-0.27 \\ g(0.75)=3^{0.75}-2=0.28 \end{gathered}[/tex]
Now, let's write the results in a table to compare them.
We can observe that an aproximate solution for f(x) = g(x) is x = -2.25.
Answer: x = -2.25.
