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Contact Joe

About me

I'm a third 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.

The Sessions:

The main aim about 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 £24 /hr
Maths A Level £24 /hr
Further Mathematics GCSE £22 /hr
Maths GCSE £22 /hr

Qualifications

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

CRB/DBS Standard

No

CRB/DBS Enhanced

No

General Availability

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Please get in touch for more detailed availability

Ratings and reviews

5from 17 customer reviews

Charlie (Student) November 2 2016

very good, clearly explains everything, makes it very easy to understand.

Charlie (Student) October 26 2016

Very good, really informative and clear. Good explanations of the chain rule for differentiation.

Barinov (Student) October 21 2016

A very good explanations

Carole (Parent) October 19 2016

My son felt it was a great lesson, very clear and well prepared for. Very happy with the first lesson. I like the way Joe checks my son's understanding all the way through the lesson.
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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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3 months ago

105 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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3 months ago

101 views
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