Contact Joe

About me

SUMMER SESSIONS: Want to boost your knowledge and ensure you are in the perfect position to start the new academic year? Your mathematical confidence and familiarity will be high and we will work towards achieving the grade that you aspire to! A-level or GCSE, we can go over whatever content you find most difficult and go through exam technique and strategy in preparation for the year ahead. I will be very flexible in this period and available for as many of these sessions as you want.

If this sounds like something you would be interested in, get in touch!

I'm a fourth year student at Bristol University studying a Masters Degree in Mathematics. I try to make my tutorials engaging, and tailor them to your individual needs so that we are working towards the grade you aspire to achieve! I've had experience tutoring Maths at GCSE and A-level standard, working with students at different levels of understanding.

About my sessions

The main aim of these sessions is to provide you with the skills and confidence to overcome the parts of maths that you find difficult. Because of this you will choose the areas that we’re going to cover in each session and I will provide the necessary materials to facilitate. What Next? If you have any questions, please book a 'Meet the Tutor Session'! Then you can tell me your exam board, what modules you are studying (if you are an A-Level student) and what you find challenging. I look forward to hearing from you!

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £26 /hr
Maths A Level £26 /hr
Further Mathematics GCSE £24 /hr
Maths GCSE £24 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
Further MathematicsA-LevelA*
Business StudiesA-LevelA*
PsychologyA-LevelA
General StudiesA-LevelA*
Disclosure and Barring Service

CRB/DBS Standard

28/10/2016

CRB/DBS Enhanced

No

General Availability

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Please get in touch for more detailed availability

Ratings and reviews

5from 78 customer reviews

Charlie (Student) May 9 2017

very helpful and very clever.

Carole (Parent) May 3 2017

Genuinely excellent. Really supportive, encouraging and clear. He is making a genuine difference to my son's progress.

Huma (Parent) March 8 2017

thanks very much great lesson

Andrea (Parent) March 7 2017

Joe has been helping my daughter for a few week and it is beginning to show. She is feeling more confident and understanding the subject more. Her one to one lessons with Joe are helping.
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Questions Joe has answered

Find the exact value of the gradient of the curve y = e^(2- x)ln(3x- 2). at the point on the curve where x = 2.

To solve this problem, we must first differentiate: Identify that we are able to use the product rule as our expression is of the form y = f(x)g(x) where f(x) = e^(2- x) and g(x) = ln(3x- 2).  Hence f'(x) = -e^(2- x) and g'(x) = 3/(3x- 2) By the product rule, dy/dx = f(x)g'(x) + f'(x)g(x) = 3...

To solve this problem, we must first differentiate:

Identify that we are able to use the product rule as our expression is of the form y = f(x)g(x) where f(x) = e^(2- x) and g(x) = ln(3x- 2). 

Hence f'(x) = -e^(2- x) and g'(x) = 3/(3x- 2)

By the product rule, dy/dx = f(x)g'(x) + f'(x)g(x) = 3e^(2- x)/(3x- 2) - e^(2- x)ln(3x- 2).

When we substitute x = 2 into this equation, we get that dy/dx = 3/4 - ln(4), which is our final answer.

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9 months ago

481 views

Solve the inequality x < 4 - |2x + 1|.

In order to deal with the modulus sign, we must take account of 2 possible cases: Case 1: |2x + 1| = (2x +1). In this case we can solve algebraicly, preserving the inequality sign, to get that x < 4 -(2x + 1) = 3 - 2x. Then by adding 2x to each side and dividing both sides by 3 we get x < 1. ...

In order to deal with the modulus sign, we must take account of 2 possible cases:

Case 1: |2x + 1| = (2x +1). In this case we can solve algebraicly, preserving the inequality sign, to get that x < 4 -(2x + 1) = 3 - 2x. Then by adding 2x to each side and dividing both sides by 3 we get x < 1.

Case 2: |2x + 1| = -(2x +1). In this case we solve algebraicly again so that x < 4 + (2x +1) = 2x + 5. Hence by subtracting a 5 and an x from each side we get x > -5.

Finally we combine the results of each case, namely that x < 1 and x > -5 to get -5 < x < 1 as our final solution.

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9 months ago

427 views
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