Functions

a) Substituting x=10 into f(x), we find,

 

f(10) = \dfrac{10}{3(10)-5} = \dfrac{10}{25}= \dfrac{2}{5}=0.4

 

b) Substituting x=2 into f(x), we find,

 

f(10) = \dfrac{10}{3(2)-5} = \dfrac{10}{1}= 10

 

c) Substituting x=-1 into f(x), we find,

 

f(10) = \dfrac{10}{3(-1)-5} = \dfrac{10}{-8}=-\dfrac{5}{4} =1.25

a) Substituting x=4 into g(x), then substituting the result into f(x),

 

g(4) = (2\times 4) - 5 = 8 - 5 = 3

 

fg(4) = f(3) = \dfrac{15}{3} = 5

 

b) For gf(-30) we must first find f(-30) and then substitute the result into g(x),

 

f(-30) = \dfrac{15}{-30} = -\dfrac{1}{2}

 

gf(-30) =  g(-\dfrac{1}{2}) = 2(-\dfrac{1}{2}) - 5 = -1 - 5 = -6

 

c) To find an expression for gf(x), substitute f(x) in for every instance of x in g(x),

 

gf(x) = 2(f(x)) - 5 = 2\times(\dfrac{15}{x}) - 5 = \dfrac{30}{x} - 5

So, we need to write the function as y=\frac{5}{x-4} and rearrange this equation to make x the subject. Then, we will swap every y with an x – and vice versa.

 

We won’t be able to get x on its own whilst it’s in the denominator, so our first step will be multiplying both sides by (x-4):

 

y(x-4)=5

 

Then, divide both sides by y:

 

x-4=\dfrac{5}{y}

 

Finally, add 4 to both sides to make x the subject:

 

x=\dfrac{5}{y}+4

 

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

 

f^{-1}(x)=\dfrac{5}{x}+4

So, we need to write the function as g=\frac{4}{x}+3 and rearrange this equation to make x the subject. Then, we will swap every g with an x – and vice versa.

 

The first step is to subtract 3 from both sides,

 

g-3=\dfrac{4}{x}+\cancel{3}-\cancel{3}

 

Then, multiply both sides by x:

 

x(g-3)=4

 

Finally, divide both sides by (g-3) to make x the subject:

 

x=\dfrac{4}{g-3}

 

Now, simply swap each x with a g and vice versa to get,

 

g^{-1}(x)=\dfrac{4}{x-3}