Hi, I’m Ryan, a second year mathematician at Durham University and I have a drive and passion to become a tutor in mathematics. I have two years of experience tutoring a group of AS level maths students so I feel I have the understanding and patience to help you. Sessions will be suited to each individual student and through explanations, examples and past paper questions we can work together to tackle specific topics you want to improve on and master. I have plenty of knowledge on all modules at A level maths and further maths. Also, I can help with students taking the STEP examinations. Feel free to get in touch and ask any questions through ‘Webmail’ or book a free ‘Meet the Tutor’ session.

Hi, I’m Ryan, a second year mathematician at Durham University and I have a drive and passion to become a tutor in mathematics. I have two years of experience tutoring a group of AS level maths students so I feel I have the understanding and patience to help you. Sessions will be suited to each individual student and through explanations, examples and past paper questions we can work together to tackle specific topics you want to improve on and master. I have plenty of knowledge on all modules at A level maths and further maths. Also, I can help with students taking the STEP examinations. Feel free to get in touch and ask any questions through ‘Webmail’ or book a free ‘Meet the Tutor’ session.

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

No DBS Check

So you are asked to find the inverse of a function f(x).

The inverse function is denoted by f ^{-1}(x).

To help with this we can use the identity f(f ^{-1}(x))=x.

Now, we need to define y=f ^{-1}(x).

Example:

f(x)=2x+1

x=f(f ^{-1}(x))=f(y)=2y+1 As f(y) is similar to f(x) but with the variable change of x to y

Hence, we need to solve:

x=2y+1

x-1=2y Minus 1 from each side of the equation

½(x-1)=y=f ^{-1}(x) As we defined f ^{-1}(x)=y

Therefore, we have found the inverse function: f ^{-1}(x) = ½(x-1)

We can continue further and find the domain and range of an inverse function using the identities:

Domain f(x) = Range f ^{-1}(x)

Range f(x) = Domain f ^{-1}(x)

So you are asked to find the inverse of a function f(x).

The inverse function is denoted by f ^{-1}(x).

To help with this we can use the identity f(f ^{-1}(x))=x.

Now, we need to define y=f ^{-1}(x).

Example:

f(x)=2x+1

x=f(f ^{-1}(x))=f(y)=2y+1 As f(y) is similar to f(x) but with the variable change of x to y

Hence, we need to solve:

x=2y+1

x-1=2y Minus 1 from each side of the equation

½(x-1)=y=f ^{-1}(x) As we defined f ^{-1}(x)=y

Therefore, we have found the inverse function: f ^{-1}(x) = ½(x-1)

We can continue further and find the domain and range of an inverse function using the identities:

Domain f(x) = Range f ^{-1}(x)

Range f(x) = Domain f ^{-1}(x)

Integration by parts formula: ∫ u*dv/dx = uv - ∫ du/dx*v dx

To solve this problem we need to use a trick by thinking of lnx as lnx*1

So we can choose: u=lnx, dv/dx=1

The next step is to find du/dx and v.

du/dx=1/x As we have differentiated each side with respect to x

v=x By integrating each side with respect to x

Now we have all the required parts to use the integration by parts formula.

∫ lnx = lnx*x – ∫ 1/x*x dx

= xlnx – ∫ 1 dx

= xlnx – x + c

Integration by parts formula: ∫ u*dv/dx = uv - ∫ du/dx*v dx

To solve this problem we need to use a trick by thinking of lnx as lnx*1

So we can choose: u=lnx, dv/dx=1

The next step is to find du/dx and v.

du/dx=1/x As we have differentiated each side with respect to x

v=x By integrating each side with respect to x

Now we have all the required parts to use the integration by parts formula.

∫ lnx = lnx*x – ∫ 1/x*x dx

= xlnx – ∫ 1 dx

= xlnx – x + c

An arithmetic sequence has a constant difference between each term.

For example: 2,4,6,8,10,12,…

We can see clearly that all the terms differ by +2.

We call this the common difference, d.

A geometric sequence has a constant ratio (multiplier) between each term.

An example is: 2,4,8,16,32,…

So to find the next term in the sequence we would multiply the previous term by 2.

This is called the common ratio, r.

These sequences are closely related as they both have the same first term, but I hope you can see how different they become if they have a common difference or a common ratio.

We can create a decreasing arithmetic sequence by choosing a negative common difference.

Similarly, a decreasing geometric sequence would have a common ratio of less than 1.

An arithmetic sequence has a constant difference between each term.

For example: 2,4,6,8,10,12,…

We can see clearly that all the terms differ by +2.

We call this the common difference, d.

A geometric sequence has a constant ratio (multiplier) between each term.

An example is: 2,4,8,16,32,…

So to find the next term in the sequence we would multiply the previous term by 2.

This is called the common ratio, r.

These sequences are closely related as they both have the same first term, but I hope you can see how different they become if they have a common difference or a common ratio.

We can create a decreasing arithmetic sequence by choosing a negative common difference.

Similarly, a decreasing geometric sequence would have a common ratio of less than 1.