# Function analysis

Hello,

given is the function $$h(y_s) = \ln (1-y_s) - \ln y_s - \gamma + \frac{\gamma}{\theta + \beta (1-y_s)}$$
my job is now to show that $h'(y_s) < 0, \forall y_s \in ]0,1[$ when
$$\frac{\gamma \beta}{\theta (\beta + \theta)} < 4$$

I guess that all constants can be assumed to be real and positive.
My first tought was to introduce the new variable $z=1-y_s$ so that after differentiation i would get

$$- 1/z + 1/(z-1) + \frac{\gamma \beta}{(\theta + \beta z)^2} < 0$$
but i cant derive the above expression from this inequality. Any help would be greatly appreciated.

Thank you.

HallsofIvy
Homework Helper
Since $0< y_s< 1[/tex], then [itex]0< z< 1$
Adding the first two terms, what you have is
$$\frac{-1}{z(z-1)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}$$
Now what is the smallest that first term could be? What is the largest the second term could be? (Remember the condition that
$$\frac{\gamma \beta}{\theta (\beta + \theta)} < 4$$).

The second term is largest when $z \rightarrow 0$, where it takes the values $$\frac{\gamma \beta}{\theta^2}$$. But as i see it, the first term is a problem since it goes toward $\pm \infty$ (or undef.?) when $z \rightarrow 0 \ \wedge \ z \rightarrow 1$ respectively.? So can this term be bounded in the interval $0<z<1$ ?

Also i believe your first term has the wrong sign.

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HallsofIvy
Homework Helper
Sorry about the sign. When I did it I was thinking 1/(1-z)- 1/z but then used your "reversed" z-1. Yes, it is
$$\frac{1}{z(z-1)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}$$
or, what I really intended,
$$\frac{-1}{z(1-z)}+\frac{\gamma\beta}{(\theta+ \beta z)^2}$$
That first term, whether 1/(z(z-1)) or -1/(z(1-z)) is always negative. And when I said "smallest value", I should have made it clear that I was thinking of the absolute value because I was focusing on the subtraction- the largest possible value of the second term, minus the smallest possible value of the first is what?

The largest possible value of the second term, minus the (abs.) smallest possible value of the first is
$$\frac{\gamma \beta}{\theta^2} - 4 < 0$$ because $$\vline \ \max_{0<z<1} \ \frac{1}{z(z-1)} \ \vline = 4$$

which leads to $$\frac{\gamma \beta}{\theta^2} < 4$$ and not the expected $$\frac{\gamma \beta}{\theta (\beta + \theta)} < 4$$

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=> or is this to be understood as a less strict requirement, because $$\frac{\gamma \beta}{\theta (\beta + \theta)} < \frac{\gamma \beta}{\theta^2}$$ ? Therefore if $$\frac{\gamma \beta}{\theta^2} < 4$$ then also $$\frac{\gamma \beta}{\theta (\beta + \theta)} < 4$$ ?