Last time, we practice some questions on functions. We will be going a step further by practicing these few examples. These questions are somewhat simple, the title is just a ''window dressing''.

Question 1

Given that $g(x)=\frac{2}{x-2}-\frac{6}{2x^2-5x+2}$, Evaluate $g^{-1}(x)$, hence solve $g^{-1}(x)=4$


Before we evaluate $g^{-1}(x)$, let first simplify $g(x)$


$2x^2-5x+2$ can be factorized into $(x-2)(2x-1)$, Accordingly 


Let subtract this using the idea of subtraction of algebraic fractions






Having simplified $g(x)$, let find $g^{-1}(x)$



Like i explain before, we switch $x$ and $y$








If you are familiar with inverse function, you can refer to this post

Having evaluated the inverse of $g(x)$, let turn our attention to $g^{-1}(x)=4$








Question 2

Given that $f(x)=\frac{1}{x+2}+\frac{2x+11}{2x^2+x-6}$, evaluate $f^{-1}(x)$


This is similar to get the first question, let simplify $f(x)$








Now, let solve for the inverse 


As usual, let switch $x$ and $y$








Question 3

If $f(x)=4-x^2$, $g(x)=\frac{5x}{2x-1}$


i) $f[g^{-1}(3)]$

ii) $g^{-1}[f(x)]=\frac{7}{5}$


First, let find the inverse of $g(x)$



Switching $x$ and $y$









Now, let solve $g^{-1}(3)$ using the idea of evaluation of function




Now, inserting $3$ in $f[g^{-1}(3)]$




ii) from (i), we see that $g^{-1}=\frac{x}{2x-5}$, Hence




Now, let equate $g^{-1}[f(x)]$ to $\frac{7}{5}$


Cross multiply










Question 4

Given that $g(x)=4-3x$ and $f(x)=x^2+ax+b$, Given also that $f[g(2)]=-5$ and $g[f(2)]=-29$, Find the value of $a$ and $b$


First, we start with $f[g(2)]=-5$




Now, we substitute $-2$ for $g(2)$ in $f[g(2)]$



Now, we equate the output of $f(-2)$ to $5$


$9=2a-b$.......eqn 1

Now, let evaluate $g[f(2)]=-29$



Now, we insert the value of $f(2)$ in $g[f(2)]$




Equating the output of $g[f(2)]$ to $-29$



$21=6a+3b$.........eqn 2

Now, let solve the equations simultaneously



We multiply Eqn 1 by 3 and eqn 2 by 1





When we subtract both eqn




To obtain $a$, we substitute the value of $b$ in Eqn 1








Help us grow our readership by sharing this post

Related Posts

Post a Comment

Subscribe Our Newsletter