Question

Consider functions fand gUsing a table of values, what is the approximate solution to the equation f(I) = g(I| to the nearest quarter of a unit?

Answer 1

Given:

Two functions f(x) and g(x) are given.

$\begin{gathered} f(x)=\frac{x-1}{x^2+x-1} \\ g(x)=3^x-2 \end{gathered}$

Required:

Find the approximate solution to the equation f(x)=g(x) to the nearest quarter of the unit.

Explanation:

(a) substitute x= 0.50in f(x) and g(x).

$\begin{gathered} f(x)=\frac{0.5-1}{(0.5)^2+0.5-1} \\ f(x)=\frac{-0.5}{-0.25} \\ f(x)=2 \end{gathered}$$\begin{gathered} g(x)=3^{0.5}-2 \\ g(x)=-0.268 \end{gathered}$

(b) substitute x=-1.75 in f(x) and g(x).

$\begin{gathered} f(x)=\frac{-1.75-1}{(-1.75)^2-1.75-1} \\ f(x)=\frac{-2.75}{0.38125} \\ f(x)=-7.2 \end{gathered}$$\begin{gathered} g(x)=3^{-1.75}-2 \\ g(x)=-1.854 \end{gathered}$

(c) substitute x= 0.75 in f(x) and g(x).

$\begin{gathered} f(x)=\frac{0.75-1}{(0.75)^2+0.75-1} \\ f(x)=\frac{-0.25}{0.3125} \\ f(x)=-0.8 \end{gathered}$$\begin{gathered} g(x)=3^{0.75}-2 \\ g(x)=2.795 \end{gathered}$

(d) substitute x= -2.25 in f(x) and g(x).

$\begin{gathered} f(x)=\frac{-2.25-1}{(-2.25)^2-2.25-1} \\ f(x)=\frac{-3.25}{1.8125} \\ f(x)=-1.793 \end{gathered}$$\begin{gathered} g(x)=3^{-2.25}-2 \\ g(x)=-1.916 \end{gathered}$

we can observe that at x = -2.25 the functions f(x) and g(x) has the approximate values.

Final Answ