KILLER QUESTIONS ON FUNCTIONS

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 $g(x)$

$g(x)=\frac{2}{x-2}-\frac{6}{2x^2-5x+2}$

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

$g(x)=\frac{2}{x-2}-\frac{6}{(x-2)(2x-1)}$.

Let subtract this using the idea of subtraction of algebraic fractions

$g(x)=\frac{2(2x-1)-6}{(x-2)(2x-1)}$

$g(x)=\frac{4x-2-6}{(x-2)(2x-1)}$

$g(x)=\frac{4x-8}{(x-2)(2x-1)}$

$g(x)=\frac{4\require{cancel}\bcancel{(x-2)}}{\require{cancel}\bcancel{(x-2)}(2x-1)}$

$g(x)=\frac{4}{2x-1}$

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

$g(x)=\frac{4}{2x-1}$

$y=\frac{4}{2x-1}$

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

$x=\frac{4}{2y-1}$

$x(2y-1)=4$

$2xy-x=4$

$2xy=4+x$

$\frac{y(2x)}{2x}=\frac{4+x}{2x}$

$y=\frac{4+x}{2x}$

$g^{-1}(x)=\frac{4+x}{2x}$

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 $g^{-1}(x)=4$

$\frac{4+x}{2x}=4$

$4+x=2x(4)$

$4+x=8x$

$4=8x-x$

$4=7x$

$\frac{4}{7}=\frac{7x}{7}$

$x=\frac{4}{7}$

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 $f(x)$

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

$f(x)=\frac{1}{x+2}+\frac{2x+11}{(2x-3)(x+2)}$

$f(x)=\frac{1(2x-3)+2x+11}{(2x-3)(x+2)}$

$f(x)=\frac{2x-3+2x+11}{(2x-3)(x+2)}$

$f(x)=\frac{4x+8}{(2x-3)(x+2)}$

$f(x)=\frac{4\require{cancel}\bcancel{(x+2)}}{(2x-3)\require{cancel}\bcancel{(x+2)}}$

$f(x)=\frac{4}{2x-3}$

Now, let solve for the inverse

$y=\frac{4}{2x-3}$

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

$x=\frac{4}{2y-3}$

$x(2y-3)=4$

$2xy-3x=4$

$2xy=4+3x$

$\frac{2x(y)}{2x}=\frac{4+3x}{2x}$

$y=\frac{4+3x}{2x}$

$f^{-1}(x)=\frac{4+3x}{2x}$

Question 3

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

Evaluate

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

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

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

$g(x)=\frac{5x}{2x-1}$

$y=\frac{5x}{2x-1}$

Switching $x$ and $y$

$x=\frac{5y}{2y-1}$

$x(2y-1)=5y$

$2xy-x=5y$

$2xy-5y=x$

$y(2x-5)=x$

$\frac{y(2x-5)}{2x-5}=\frac{x}{2x-5}$

$y=\frac{x}{2x-5}$

$g^{-1}(x)=\frac{x}{2x-5}$

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

$g^{-1}(3)=\frac{3}{2(3)-5}$

$g^{-1}(3)=\frac{3}{1}$

$g^{-1}(3)=3$

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

$f(3)=4-(3)^2$

$f(3)=4-9$

$f(3)=-5$

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

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

$g^{-1}[f(x)]=\frac{4-x^2}{8-2x^2-5}$

$g^{-1}[f(x)]=\frac{4-x^2}{3-2x^2}$

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

$\frac{4-x^2}{3-2x^2}=\frac{7}{5}$

Cross multiply

$5(4-x^2)=7(3-2x^2)$

$20-5x^2=21-14x^2$

$-5x^2+14x^2=21-20$

$9x^2=1$

$\frac{9x^2}{9}=\frac{1}{9}$

$x^2=\frac{1}{9}$

$\sqrt{x^2}=\sqrt{\frac{1}{9}}$

$x=\frac{1}{3}$

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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 $f[g(2)]=-5$

$g(2)=4-3(2)$

$g(2)=4-6$

$g(2)=-2$

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

$f(-2)=(-2)^2+(-2)a+b$

$f(-2)=4-2a+b$

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

$4-2a+b=-5$

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

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

$f(2)=(2)^2+a(2)+b$

$f(2)=4+2a+b$

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

$g[f(2)]=4-3(4+2a+b)$

$g[f(2)]=4-12+6a+3b$

$g[f(2)]=-8+6a+3b$

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

$-8+6a+3b=-29$

$+29-8=6a+3b$

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

Now, let solve the equations simultaneously

$9=2a-b$

$21=6a+3b$

We multiply Eqn 1 by 3 and eqn 2 by 1

$3(9=2a-b)$

$1(21=6a+3b)$

$27=6a-3b$

$21=6a+3b$

When we subtract both eqn

$6=-6b$

$\frac{6}{-6}=-\frac{-6b}{-6}$

$b=-1$

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

$9=2a-(-1)$

$9=2a+1$

$9-1=2a$

$8=2a$

$\frac{8}{2}=\frac{2a}{2}$

$4=a$

$a=4$

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