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 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.
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!
|Further Mathematics||A Level||£24 /hr|
|Maths||A Level||£24 /hr|
|Further Mathematics||GCSE||£22 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
Please get in touch for more detailed availability
Charlie (Student) November 2 2016
Charlie (Student) October 26 2016
Barinov (Student) October 21 2016
Carole (Parent) October 19 2016
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.see more
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.see more