Melody

Thanks David 

Alan's answer is good but I'll give you another answer Stu - I got called away before.  Sorry.

The LaTex fraction display is not  working properly so my presentation is not a good as I wanted.

$$f(x)=x-2ln(x+1)\:\:\: 0\le x\le2\\

f'(x)=1- 2/(x+1)\\

f'(x)=1-2(x+1)^{-1}\\

f''(x)=+2(x+1)^{-2}\\

f''(x)=2/(x+1)^2$$

Since the (x+1) is squared, the second derivative is always positive and this means that any turning point will be a minimum.  Which in turn means that the two end points will be higher.

Now we need to find any turning points   f'(x)=0

$$0=1-2/(x+1)\\

2/(x+1)=1\\

2 = x+1\\

x=1\\

f(1)=1-2ln(1+1) = 1-2ln2 \approx -0.386$$

So (1,-0.386) is the minumum

Now to find the maximum - you need to look at the end points.

f(0)=0-2ln(0+1) = 0

f(2)=2-2ln(2+1) = 2-2ln3 = -0.197

So the maximum is 0 at x=0 and the minimum is 1-2ln2 at x=1

I don't know how standard it is - It is what I was taught and so it is what I teach.

 (I was lucky, I had a couple of really good maths teachers when I was at school, most of what they taught me was correct.  Now I pass that learning on.  It may not all be flawless) 

Hi bioschip,

I just thought I would add a litte

2 angles that add to 90 degrees are called complementary angles

2 angles that add to 180 degrees are called supplementary angles.

If you can remember the 2 names it is easy to remember which is which because

2 90s make 180 and

2 c's make an s (one is back to front)

does that make sense?

Hi Stu,

It is nice to see you again.

Is there supposed to be a - sign in front of the f(x) or is that just a dash?

$$\frac{d}{dx}x=1$$

In other words the gradient of the line y=x is 1

I don't get it.  

Thanks Rom,

I'd like to get back and have a proper look at this one!   

hi Professor,  It is great that you answered this question.  I love people to get involved.

Please edit your last step to make it correct, I am reluctant to give you points until it is fixed.   

What do you mean.

If you want inverse tan it is just 2nd tan

It may help if I tell you that inverse tan is the same as arc tan

tan^-1 = atan