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A goalkeeper kicked the ball such that the ball describes a given curve, until it arrived the ground again. The curve is defined by $h(x)=2x+10\ln(1-0.1x)$ where $x$ is the distance between the point where the goalkeeper kicked the ball and the ball's projection on the ground while it describes the curve on the air. The function $h$ gives the height where the ball is as $x$ increases.

What is the maximum distance that the ball can reach?

My interpretation of this question was: In what interval of $x$ values the function makes sense? Well, in this case the height is a positive number, so in order to this problem have sense $h$ should be positive too. I tried to define the domain.Since the distance, $x$ is positive and there is a $\ln$:

$(x\geq0) \wedge (1-0.1x\gt 0)$

The domain's interval is $[0,10[$. But at values of $x$ near $10$ the function $h$ becomes negative.Then I thought, what about finding the roots of the function?

One I allready know $(x=0)$, but I don't know how to find the other.

Help me to solve $h(x)=0$. Thanks

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    The question is indeed: What's the interval of x values for what the function makes sense?If the distance, x and the height, h have to be positive.The function only makes sense on the right of y axis and above the x axis.In other words: in the first quadrant.2011-11-25

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You are asking for the other positive root of the equation $f(x)=0$, where $f(x)=2x+\ln(1-0.1x ).$ The solution cannot be expressed exactly by using simple combinations of familiar functions and constants. But if we think of the problem as a practical problem, we can obtain an answer that is as accurate as we need.

There are many possible approaches. For example,if you have access to a graphing calculator, you can ask it to graph the curve $y=f(x)$. Then you can read the second root off the display. To get additional accuracy, use the "zoom" feature.

Various graphing programs are available, for a wide variety of computer platforms. This includes a number of free programs.

Wolfram Alpha can be freely used. It will graph for you, solve equations for you, and much much more.

Finally, we can attack the problem ourselves. It is useful to have access to a simple scientific calculator, or to a calculator program. (Such a program is probably already on the computer that you use.)

Many root-finding algorithms have been developed. A useful one to know is the Newton-Raphson Method. An often even better one is the Secant Method. Once these methods get a good smell of the root, they usually home in on it with remarkable efficiency.

We will use a less fancy but still systematic method, often called the Bisection Method. It will be best if you do the calculations below with your own calculator.

We need two starting points, one of them smaller than the root, and the other bigger than the root. Preferably these starting points are close to each other.

Our curve $y=2x+10\ln(1-0.1x)$ goes up and then down. Have we hit the ground by the time $x=7$? Let's calculate $f(7)$. No, $f(7)$ is about $1.96$. Now let's calculate $f(9)$. We get $f(9)\approx -5.0$.

So far we have $f(7)>0$, and $f(9)<0$. Since our function $f$ is continuous, there is a root between $x=7$ and $x=9$.

Now look at the point halfway between $7$ and $9$, namely $8$. So we bisect the interval. Calculate $f(8)$. We get $-0.09$. So $x=8$ is awfully close to a root. Not a very impressive kick!

Now I will be a little silly, and continue like a machine. We have $f(7)>0$, and $f(8)<0$, so look at the point halfway between, namely $7.5$. We have $f(7.5)>0$. Since $f(8)<0$, our root is between $7.5$ and $8$. Look at the point halfway in between, namely $7.75$.

We have $f(7.75)>0$, so our root is between $7.75$ and $8$. Look at the point halfway in between, namely $7.875$. We find that $f(7.875)>0$. So there is a root between $7.875$ and $8$. Go halfway in between, to $7.9375$. We have $f(7.9375)>0$. Go halfway between $7.9375$ and $8$, to $7.96875$. We find that $f(7.96875)<0$. So our root is between $7.9375$ and $7.96875$.

If we need greater accuracy, continue. Each time we go through one cycle of the Bisection Method, our region of uncertainty shrinks by a factor of $2$. After a while, we are close enough for all practical purposes.

Remark: In our case, since $x=8$ is so close to a root, you may have been just intended to just play around for a bit, and notice that $f(8)$ is nearly $0$.

You might want to look up the Newton Method, or the Secant Method, and see how they perform.

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    Thanks for the answer.It was ver$y$ clarif$y$.2011-11-26
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The solution below is based on the fact that you have high-school level of math.

You have correctly identified the range to be between 0 and 10.

in the hunt for x, we examine the h(x) in the interval of the range end points looking for the sing of h(x). In any interval, if h(x) changes it is sign, that would indicate that a root exists in that interval. So let's try this in [0.1,10] (I did not choose zero because we know that there is a root at zero, and that it is not the one we are looking for).

you will find that h(0.1) > 0 and h(10) < 0, so we can divide this interval into 2 intervals:

[0.1,5] and [5.1,10]

h(0.1) and h(5) are both positive whereas h(5.1) > 0 and h(10) < 0 this indicates that the value we want must be in [5.1, 10]. Let's split this range again into 2 intervals:

[5.1,7.5] and [7.6,10]

applying the same logic as above, you can exclude [5.1,7.5] since h(x) > 0 at both ends. This leaves us with [7.6,10] at which the desired root lies.

You can repeat this method as many times until you get h(x) close to zero as you desire. You should eventually get the value x=7.9 as suggested by @Arturo Magidin. The more iterations you make, the more accurate your approximation will be.

The above steps roughly follows what is called the Bisection Method for finding roots.

The picture below shows how the intervals halving progressively can narrow down the interval where the root lies.Interval Halving

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Have you studied Newton's Method for numerical (that is, approximate) solution of equations?

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    The Workbook and the teacher allows the use of the graph calculator whitout the proof.2011-11-26