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Degree: Natural Sciences with Year Abroad (Bachelors) - Durham University
My name is Anna Hutchinson and I am 21 years old. I am studying Natural Sciences (Maths and Biology) at Durham University, but I am currently on my year abroad in Canada, studying Mathematics at Calgary University - but don't worry I can work around the time-zone!
I have tutored in the past, both at school and University and I find it easy to adapt my teaching style in order to suit the student. I understand it may be hard to grasp particular topics and this can get frustrating! As a result I always have a few tricks up my sleeve to help make the topic seem a bit more straight forward.
After getting two A*s and an A in my A-Levels, and a very high 2:1 in my second year of university, I thoroughly understand the subjects I am offering and my passion for them often spreads to my tutees.
I am very friendly and approachable and would love to help you (or your child) to not only improve their grades, but to understand the topics better as a whole.
Feel free to book a 'meet the tutor' session to get to know me a bit better and I look forward to meeting you :)
|Biology||A Level||£22 /hr|
|Maths||A Level||£22 /hr|
|Maths||13 Plus||£20 /hr|
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John (Student) March 21 2015
A stationary point is simply a point on a graph where the derivative=0. Ie, the rate of change of the curve at this point is 0 and therefore it is neither increasing or decreasing at this point.
There are three types you need to know about:
1) A maximum: Here the derivative =0 and the second derivative <0.
2) A minimum: Here the derivative =0 and the second derivative >0
3) A point of inflection: Here the derivative and the second derivative =0
Note, the second derivative means the derivative of the first derivative!
If at a point, say c, f'(c)=0 then there is a stationary point at this value of x.
Differentiate f'(x) to get the second derivative.
Plug in the value of c again and if the solution is..
0 - Point of inflection
Positive - Minimum turning point
Negative - Maximum turning point
y = x3 - 6x2 + 9x - 4
Find any stationary points and determine their nature.
dy/dx = 3x2- 12x + 9
At a stationary point, dy/dx=0
So 3x2- 12x + 9 = 0
3(x2- 4x + 3) = 0
(x - 3)(x - 1) = 0
So stationary point at x = 3 and x = 1.
Now, to determine the nature of these..
f''(x) = 6x - 12
f''(3) = 18 - 12 = 6 therefore minimum turning point at x = 3
f''(1) = 6 - 12 = -6 therefore maximum turning point at x = 1see more