Global max-min

Incidentally, to import the graph I used the Tiny Pic upload service now available with the imported image facility (icon on the far right of the formatting ribbon).  I created the graph in a piece of software on my PC; used the Windows snipping tool to copy and save it as a .png file; uploaded it using the Tiny Pic option in insert image; copied the url and pasted that into the Image URL box in insert image.  

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?

Assuming this is f(x) = x - 2ln(x+1), with 0≤x≤2, the best thing to do first is to plot a graph to see what it looks like.  In this case you will see a curve with a single minimum.

Differentiate the function with respect to x and set the result to zero. 

$$1-\frac{2}{x+1}=0$$

This will give x = 1.  So the minimum occurs at x = 1 and the function has a value of f(1) = 1 - 2ln(2) there.

$${\mathtt{1}}{\mathtt{\,-\,}}{\mathtt{2}}{\mathtt{\,\times\,}}{ln}{\left({\mathtt{2}}\right)} = -{\mathtt{0.386\: \!294\: \!361\: \!119\: \!890\: \!6}}$$

I dont get it. But I need to find the global minima and global maxima for this equation. What was the minima and x value, and same for maxima? 

What steps are used to get the answers?

1.  Look at a graph of your function between the limits of 0 and 2:

You can see that there is a single minimum in the range.

2. To find the turning points on a curve you need to find the places where the tangents are zero.  A zero tangent means that a line through that point is horizontal.  You can see from the curve that a tangent line at the minimum of the curve will be horizontal.

3. The tangent line to a curve at any point has the same gradient as the curve at that point.  The gradient is obtained by differentiating the function with respect to x. So if the curve is described by f(x) = x - 2ln(x+1) you have to differentiate this to find the gradient.  Differentiating the first term, x, with respect to x is just 1.  Differentiating the second term with respect to x gives 2/(x+1).  Overall then the gradient of the curve at any point is given by gradient(x) = 1-2/(x+1).

4.  We want the point where this gradient is zero, so we set 1-2/(x+1) = 0 and find the value of x that makes this true.  Here we can see that x = 1 is required since 1-2/(1+1) is indeed zero.

5.  The value of f(x) at this point is obtained by putting x = 1 into the expression for f(x) -  this will give us the value of the minimum of f(x):  1-2ln(1+1)  or 1-2ln(2) = -0.386...

6.  You can see that there isn't a turning point maximum in the range from x = 0 to x = 2, but the maximum value of the function in this range clearly happens when x = 0.  The value of the function here is f(0) = 0 - 2ln(0+1) = -2ln(1) = 0.

Hope this helps a bit more. 

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