# 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 $f[g(x)]=\frac{4}{9}$

$f[g(x)]=\frac{4}{9}$ 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

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

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

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

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

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

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

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

$2x=-1$ 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)-3f(1)=1-2-3f(1)=-4$Next, we substitute$-4$for$f(1)$in$g[f(1)]g(-4)=a(-4)^2+2g(-4)=16a+2$Having gotten the output value of$g(-4)$, let equate it to$616a+2=616a=6-216a=4\frac{16a}{16}=\frac{4}{16}a=\frac{1}{4}$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$g(x)g(x)=\frac{2}{x-3}+3g(x)=\frac{2+3(x-3)}{x-3}g(x)=\frac{2+3x-9}{x-3}g(x)=\frac{3x-7}{x-3}$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$yx=\frac{3y-7}{y-3}x(y-3)=3y-7xy-3x=3y-7xy-3y=-7+3x\frac{y(x-3)}{x-3}=\frac{3x-7}{x-3}y=\frac{3x-7}{x-3}g^{-1}=\frac{3x-7}{x-3}$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$yx=2y-1x+1=2yy=\frac{x+1}{2}$Now, we back-substitute$f^{-1}(x)$for$yf^{-1}(x)=\frac{x+1}{2}$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$(x+1)(4x-2)=9\times2x(4x-2)+1(4x-2)=184x^2-2x+4x-2=184x^2+2x-2-18=04x^2+2x-20=0\frac{4x^2}{2}+\frac{2x}{2}-\frac{20}{2}=\frac{0}{2}2x^2+x-10=0$Solving quadratically$2x^2+5x-4x-10=0x(2x+5)-2(2x+5)=0x=2$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$f(x-1)=2(x-1)^2-1f(x-1)=2(x^2-2x+1)-1f(x-1)=2x^2-4x+2-1f(x-1)=2x^2-4x+1$Having gotten the value of$f(x-1)$, let evaluate$f(x-1)-7=02x^2-4x+1-7=02x^2-4x-6=0$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,$x^2-3x+x-3=0x(x-3)+1(x-3)=0x=-1, x=3$Therefore the values of x are -1 ans 3 ii) First, We evaluate each separately if$f(x)=2x^2-1$, then$f(\frac{-1}{2})=2(\frac{-1}{2})^2-1f(\frac{-1}{2})=2(\frac{1}{4})-1f(\frac{-1}{2})=\frac{2}{4}-1f(\frac{-1}{2})=\frac{1}{2}$if$g(x)=3x+2$, then$g(3)=3(3)+2g(3)=11$If$f(x)=2x^2-1$,then$f(4)=2(4)^2-1f(x)=32-1=31g(x)=3x+2g(5)=3(5)+2g(5)=17$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.

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