**About Me**

I am a third year mathematics student at Durham University. Friendly and patient, I look forward to helping you feel confident with any aspect of maths at GCSE and A-level. I have previous tutoring experience at both of these levels with students of a variety of abilities; whether you are pushing for a pass or aiming for an A*, I will ensure that you achieve your full potential.

**What to Expect From a Session**

Together we will set reasonable targets for each session, based on the topics *you* want to focus on. I will help build your understanding of the topic until we feel confident enough to attempt practice questions together.

Expect lots of useful diagrams, analogies and examples. You will be accommodated for, no matter what type of learner you are!

Aside from all this, I hope that I can impart some of my love for maths with you. It really is a deeply interesting subject, and I'll make sure we have fun whilst learning.

**How Do I Get in Touch?**

If you'd like to talk with me some more before booking any sessions, please do not hesitate from booking a 'Meet the Tutor' session with me. I look forward to having a chat about anything! Also feel free to send me a WebMail message with any further questions.

I look forward to meeting you!

Alex

**About Me**

I am a third year mathematics student at Durham University. Friendly and patient, I look forward to helping you feel confident with any aspect of maths at GCSE and A-level. I have previous tutoring experience at both of these levels with students of a variety of abilities; whether you are pushing for a pass or aiming for an A*, I will ensure that you achieve your full potential.

**What to Expect From a Session**

Together we will set reasonable targets for each session, based on the topics *you* want to focus on. I will help build your understanding of the topic until we feel confident enough to attempt practice questions together.

Expect lots of useful diagrams, analogies and examples. You will be accommodated for, no matter what type of learner you are!

Aside from all this, I hope that I can impart some of my love for maths with you. It really is a deeply interesting subject, and I'll make sure we have fun whilst learning.

**How Do I Get in Touch?**

If you'd like to talk with me some more before booking any sessions, please do not hesitate from booking a 'Meet the Tutor' session with me. I look forward to having a chat about anything! Also feel free to send me a WebMail message with any further questions.

I look forward to meeting you!

Alex

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

**What are the maxima and minima?**

The maxima of a function f(x) are all the points on the graph of the function which are 'local maximums'. A point where x=a is a local maximum if, when we move a small amount to the left (points with xa), the value of f(x) *decreases*. We can visualise this as our graph having the peak of a 'hill' at x=a.

Similarly, the minima of f(x) are the points for which, when we move a small amount to the left or right, the value of f(x) *increases*. We call these points 'local minimums', and we can visualise them as the bottom of a 'trough' in our graph.

One similarity between the maxima and minima of our function is that the gradient of our graph is always equal to 0 at all of these points; at the very top of the peaks and the very bottom of the troughs, the slope of our graph is completely flat. **This means our derivative, f '(x), is equal to zero at these points.**

**How do we find them?**

1) Given f(x), we differentiate once to find f '(x).

2) Set f '(x)=0 and solve for x. Using our above observation, the x values we find are the 'x-coordinates' of our maxima and minima.

3) Substitute these x-values back into f(x). This gives the corresponding 'y-coordinates' of our maxima and minima.

**Which of these points are maxima and which are minima?**

Here we may apply a simple test. Assume we've found a stationary point (a,b):

1) Differentiate f '(x) once more to give f ''(x), the *second derivative*.

2) Calculate f ''(a). If f ''(a)<0 then (a,b) is a local maximum.

If f ''(a)>0 then (a,b) is a local minimum.

To see why this works, imagine moving gradually towards our point (a,b), plotting the slope of our graph as we move. If our point is a local maximum, we can that this slope starts off positive, decreases to zero at the point, then becomes negative as we move through and past the point. Our slope, f '(x), is *decreasing* throughout this movement, so we must have that f ''(a)<0.

The exact reverse is true if (a,b) is a local minimum. Our slope is *increasing* through the same movement, so here we have that f ''(a)>0.

**An example**

**Find the maxima and minima of f(x)=x ^{3}+x^{2}.**

First, we find f '(x). Using the rules of differentiation, we find f '(x)=3x^{2}+2x.

Now let's set f '(x)=0: 3x^{2}+2x=0

x(3x+2)=0 (factorising)

**x=0** or **x=-2/3**

Substitute these values back in so that we can find our 'y-coordinates': f(0)=(0)^{3}+(0)^{2}=0

f(-2/3)=(-2/3)^{3}+(-2/3)^{2}=4/27

Hence our stationary points are **(0,0)** and **(-2/3,4/27)**.

Finally, we use our test: f ''(x)=6x+2

f ''(0)=2 (substituting x=0)

f ''(-2/3)=-2 (substituting x=-2/3)

2>0, so (0,0) is a *local minimum* of f(x).

-2<0, so (-2/3,4/27) is a *local maximum* of f(x).

**What are the maxima and minima?**

The maxima of a function f(x) are all the points on the graph of the function which are 'local maximums'. A point where x=a is a local maximum if, when we move a small amount to the left (points with xa), the value of f(x) *decreases*. We can visualise this as our graph having the peak of a 'hill' at x=a.

Similarly, the minima of f(x) are the points for which, when we move a small amount to the left or right, the value of f(x) *increases*. We call these points 'local minimums', and we can visualise them as the bottom of a 'trough' in our graph.

One similarity between the maxima and minima of our function is that the gradient of our graph is always equal to 0 at all of these points; at the very top of the peaks and the very bottom of the troughs, the slope of our graph is completely flat. **This means our derivative, f '(x), is equal to zero at these points.**

**How do we find them?**

1) Given f(x), we differentiate once to find f '(x).

2) Set f '(x)=0 and solve for x. Using our above observation, the x values we find are the 'x-coordinates' of our maxima and minima.

3) Substitute these x-values back into f(x). This gives the corresponding 'y-coordinates' of our maxima and minima.

**Which of these points are maxima and which are minima?**

Here we may apply a simple test. Assume we've found a stationary point (a,b):

1) Differentiate f '(x) once more to give f ''(x), the *second derivative*.

2) Calculate f ''(a). If f ''(a)<0 then (a,b) is a local maximum.

If f ''(a)>0 then (a,b) is a local minimum.

To see why this works, imagine moving gradually towards our point (a,b), plotting the slope of our graph as we move. If our point is a local maximum, we can that this slope starts off positive, decreases to zero at the point, then becomes negative as we move through and past the point. Our slope, f '(x), is *decreasing* throughout this movement, so we must have that f ''(a)<0.

The exact reverse is true if (a,b) is a local minimum. Our slope is *increasing* through the same movement, so here we have that f ''(a)>0.

**An example**

**Find the maxima and minima of f(x)=x ^{3}+x^{2}.**

First, we find f '(x). Using the rules of differentiation, we find f '(x)=3x^{2}+2x.

Now let's set f '(x)=0: 3x^{2}+2x=0

x(3x+2)=0 (factorising)

**x=0** or **x=-2/3**

Substitute these values back in so that we can find our 'y-coordinates': f(0)=(0)^{3}+(0)^{2}=0

f(-2/3)=(-2/3)^{3}+(-2/3)^{2}=4/27

Hence our stationary points are **(0,0)** and **(-2/3,4/27)**.

Finally, we use our test: f ''(x)=6x+2

f ''(0)=2 (substituting x=0)

f ''(-2/3)=-2 (substituting x=-2/3)

2>0, so (0,0) is a *local minimum* of f(x).

-2<0, so (-2/3,4/27) is a *local maximum* of f(x).