32. Functions (harder)

Question 1

LEVEL 8

Given that f(x) = 2x+1

Find f^{-1}(5)

Select the correct answer from the list below:

A: 2

B: -2

C: \dfrac{1}{11}

D: - \dfrac{1}{11}

CORRECT ANSWER:    A: 2

WORKED SOLUTION:

First we can swap the terms so replace x with y and y with x.

x = 2y+1

Then we can rearrange to make y the subject again.

2y=x-1

y=\dfrac{x-1}{2}

f^{-1}(x)=\dfrac{x-1}{2}

So,

f^{-1}(5)=\dfrac{5-1}{2} = 2

Question 2

LEVEL 8

f(x)=2x+4 and g(x)=5x-2

Find the value of fg(9)

Select the correct answer from the list below:

A: 90

B: 54

C: 60

D: 87

CORRECT ANSWER:   A: 90

WORKED SOLUTION:

fg(9) is the short of writing “do g(9) and the put that answer into f(x).”

fg(9)=f(g(9))

g(9)=5\times9-2=45-2=43

fg(9)=f(g(9))=f(43)=2\times43+4=86+4=90

Question 3

LEVEL 8

f(x)=\dfrac{2}{x+1} and g(x)=5x+2

Find the value of gf(x)

Select the correct answer from the list below:

A: \dfrac{17-3x}{x+1}

B: \dfrac{x+1}{2x+5}

C: \dfrac{5x+2}{x+1}

D: \dfrac{2x+12}{x+1}

CORRECT ANSWER:    D: \dfrac{2x+12}{x+1}

WORKED SOLUTION:

gf(x) is the short of writing “replace every x in g with what f(x) is equal to.”

gf(x)=g(f(x))

g(f(x))=5\times f(x)+2

g(f(x))=5\times \frac{2}{x+1}+2=\frac{10}{x+1}+2=\frac{10}{x+1}+2\frac{x+1}{x+1}=\frac{10}{x+1}+\frac{2x+2}{x+1} =\frac{10+2x+2}{x+1}=\frac{2x+12}{x+1}

Question 4

LEVEL 8

Find the inverse f^{-1}(x) of the function f(x)=2x^2+4

Select the correct answer from the list below:

A: f^{-1}(x)=\sqrt{\frac{x-4}{2}}

B: f^{-1}(x)=\frac{1}{2x^2+4}

C: f^{-1}(x)= 2x^2+3

D: f^{-1}(x)=\sqrt{2x^2+4}

CORRECT ANSWER:     A: f^{-1}(x)=\sqrt{\frac{x-4}{2}}

WORKED SOLUTION:

To start we write the function as y=2x^2+4 and rearrange to make x the subject. To rearrange, we do BIDMAS backwards.

Do we have any subtractions? No, move on.

Do we have any additions? Yes, and we do the inverse (subtraction).

y=2x^2+4

y-4=2x^2+4-4

y-4=2x^2

Do we have any multiplications? Yes, and we do the inverse (division).

y-4=2x^2

(y-4)\div2=2x^2\div2

\frac{y-4}{2}=x^2

Do we have any divisions? No, move on.

Do we have any indices? Yes, and we do the inverse (Square root here)

\frac{y-4}{2}=x^2

\sqrt{\frac{y-4}{2}}=\sqrt{x^2}

\sqrt{\frac{y-4}{2}}=x

x=\sqrt{\frac{y-4}{2}}

Now, swap each x with a y and vice versa to get:

To express this properly, we now sway the y with f^{-1}(x)

Question 5

LEVEL 8

Find the inverse g^{-1}(x) of the function g(x)=\frac{5}{2x+4}

Select the correct answer from the list below:

A: f^{-1}(x)= 1-\frac{5}{2x+4}   

B: f^{-1}(x)= 10x-4

C: f^{-1}(x)= \frac{5}{2x}-2 

D: f^{-1}(x)= \frac{2x+4}{5} 

CORRECT ANSWER:     C: f^{-1}(x)= \frac{5}{2x}-2 

WORKED SOLUTION:

To start we write the function as y=\frac{5}{2x+4} and rearrange to make x the subject. To rearrange, we do BIDMAS backwards.

Do we have any subtractions? No, move on.

Do we have any additions? No, not by itself (ours is inside the division)

Do we have any multiplications? No, not by itself (ours is inside the division)

Do we have any divisions? Yes, the fraction is a division so we do the inverse (multiply by the denominator).

y=\frac{5}{2x+4}

y\times(2x+4)=\frac{5}{2x+4}\times(2x+4)

y\times(2x+4)=5

This has now given us a multiplication, so we have to do the inverse first (divide by y)

y\times(2x+4)\div y=5\div y

2x+4=\frac{5}{y}

This has now revealed the addition, so we need to get rid of that by subtracting.

2x+4=\frac{5}{y}

2x+4-4=\frac{5}{y}-4

2x=\frac{5}{y}-4

And finally, we now have another multiplication, so we need to divide.

2x\div2=(\frac{5}{y}-4)\div2

x=\frac{5}{y}\div2-4\div2

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

Now, swap each x with a y and vice versa to get:

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

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

To express this properly, we now sway the y with f^{-1}(x)