HARD QUESTIONS ON FUNCTION

So far, we have discussed the types of function, the inverse of a function, and composite function.

Today, we will practice some questions on function. So, let's get started.

Question 1

Given that $f(x)=x^2$ and $g(x)=\frac{1}{x+2}$, Evaluate $f[g(x)]=\frac{4}{9}$

Answer

$f[g(x)]=\frac{4}{9}$ means that we should evaluate $f[g(x)]$ and then equate the output value to $\frac{4}{9}$

Now, let evaluate $f[g{x}]$

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

$f[g(x)]=(\frac{1}{x+2})(\frac{1}{x+2})$

$f[g(x)]=\frac{1}{x(x+2)+2(x+2)}$

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

Having gotten the output of $f[g(x)]$, let now equate it to $\frac{4}{9}$

$\frac{1}{x^2+4x+4}=\frac{4}{9}$

Cross multiply

$4(x^2+4x+4)=1\times9$

$4x^2+16x+16=9$

$4x^2+16x+16-9=0$

$4x^2+16x+7=0$

Now, let solve this quadratically

$4x^2+14x+2x+7=0$

$2x(2x+7)+1(2x+7)=0$

$(2x+1)(2x+7)=0$

$2x=-1$ or 2x=-7$

$\frac{2x}{2}=\frac{-1}{2}$ or $\frac{2x}{2}=\frac{-7}{2}$

$x=\frac{-1}{2}$ or $\frac{-7}{2}$

Question 2

Given that $f(x)=x^2-2x-3$ and $g(x)=ax^2+2$. Find the value of $a$ for which $g[f(1)]=6$

Answer

First, we solve for $f[1]$ using the idea of evaluation of function

$f(1)=(1)^2-2(1)-3$

$f(1)=1-2-3$

$f(1)=-4$

Next, we substitute $-4$ for $f(1)$ in $g[f(1)]$

$g(-4)=a(-4)^2+2$

$g(-4)=16a+2$

Having gotten the output value of $g(-4)$, let equate it to $6$

$16a+2=6$

$16a=6-2$

$16a=4$

$\frac{16a}{16}=\frac{4}{16}$

$a=\frac{1}{4}$

Question 3

If $f(x)=\frac{2}{x}$, $g(x)=f(x-3)+3$, Evaluate $g(x)$. Hence, prove that $g(x)$ is a self inverse function

Answer

$g(x)=f(x-3)+3$ 

First, let evaluate $f(x-3)$

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

Now, let substitute $f(x-3)$ in $g(x)$

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

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

$g(x)=\frac{2+3x-9}{x-3}$

$g(x)=\frac{3x-7}{x-3}$

Now, let prove that $g(x)$ is a self inverse function.

But, what is a self inverse function?

A self inverse is a function where the original function is the same as its inverse function. That is, a self inverse function is a function where:

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

So, let prove:

$g(x)=\frac{3x-7}{x-3}$

Now, let find the inverse function

$y=\frac{3x-7}{x-3}$

Now, let switch $x$ and $y$

$x=\frac{3y-7}{y-3}$

$x(y-3)=3y-7$

$xy-3x=3y-7$

$xy-3y=-7+3x$

$\frac{y(x-3)}{x-3}=\frac{3x-7}{x-3}$

$y=\frac{3x-7}{x-3}$

$g^{-1}=\frac{3x-7}{x-3}$

As can be seen, the inverse of the function is $\frac{3x-7}{x-3}$ which is the same as the original function. Therefore, $g(x)=\frac{3x-7}{x-3}$ is a self inverse function.

Question 4

Given that $f(x)=2x-1$, Evaluate

i) $f^{-1}(x)$

ii) $f^{-1}(x)=\frac{9}{2f(x)}$

Answer

i)$y=2x-1$

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

$x=2y-1$

$x+1=2y$

$y=\frac{x+1}{2}$

Now, we back-substitute $f^{-1}(x)$ for $y$

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

ii) $f^{-1}(x)=\frac{9}{2f(x)}$ 

From eqn i, $f^{-1}(x)=\frac{x+1}{2}$, thus:

$\frac{x+1}{2}=\frac{9}{2(2x-1)}$

$\frac{x+1}{2}=\frac{9}{4x-2}$

Cross multiply

$(x+1)(4x-2)=9\times2$

$x(4x-2)+1(4x-2)=18$

$4x^2-2x+4x-2=18$

$4x^2+2x-2-18=0$

$4x^2+2x-20=0$

$\frac{4x^2}{2}+\frac{2x}{2}-\frac{20}{2}=\frac{0}{2}$

$2x^2+x-10=0$

Solving quadratically

$2x^2+5x-4x-10=0$

$x(2x+5)-2(2x+5)=0$

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

Question 5

Two functions. F and g are defined by $f(x)=2x^2-1$ and  $g(x)=3x+2$ where x is a real number.

i.  if $f(x-1)-7=0$, find the value of x

ii. evaluate $\frac{f\left(\frac{-1}{2}\right)\times g(3)}{f(4)-g(5)}$

Solutions

I. if $f(x)=2x^2-1$, then

$f(x-1)=2(x-1)^2-1$

$f(x-1)=2(x^2-2x+1)-1$

$f(x-1)=2x^2-4x+2-1$

$f(x-1)=2x^2-4x+1$

Having gotten the value of $f(x-1)$, let evaluate $f(x-1)-7=0$

$2x^2-4x+1-7=0$

$2x^2-4x-6=0$

For simplicity, let's divide through by 2

$\frac{2x^2}{2}-\frac{4x}{2}-\frac{6}{2}=\frac{0}{2}$

$x^2-2x-3=0$

Solving quadratically,

$x^2-3x+x-3=0$

$x(x-3)+1(x-3)=0$

$x=-1, x=3$

Therefore the values of x are -1 ans 3

ii) First, We evaluate each separately

if $f(x)=2x^2-1$, then

$f(\frac{-1}{2})=2(\frac{-1}{2})^2-1$

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

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

$f(\frac{-1}{2})=\frac{1}{2}$

if $g(x)=3x+2$, then

$g(3)=3(3)+2$

$g(3)=11$

If $f(x)=2x^2-1$,then

$f(4)=2(4)^2-1$

$f(x)=32-1=31$

$g(x)=3x+2$

$g(5)=3(5)+2$

$g(5)=17$

Therefore, 

$\frac{f\left(\frac{-1}{2}\right)\times g(3)}{f(4)-g(5)}=\frac{\frac{-1}{2}\times 11}{31-17}$

$\frac{f\left(\frac{-1}{2}\right)\times g(3)}{f(4)-g(5)}=\frac{\frac{-11}{2}}{14}$

$\frac{f(\left(\frac{-1}{2}\right)\times g(3)}{f(4)-g(5)}=\frac{-11}{28}$

That is all there is to it. In our next lesson, we'll go over some examples that will help you solidify your understanding of function.

Meanwhile, if you have any questions, please ask them in our telegram community

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