# Solve this calculus questions

Let a differentiable function $f(x)$ satisfy the rule

$f(xy) = f(x) + f(y) + xy -x - y$ for all $x,y>0$ Given

$f^1(1)=4$

If $f(x_o) = 0$ then,

Find the interval in which $x_o$ lies in?

## 1 Answer

Well, I have the following idea:

Lets note that

$$f(x\cdot 1) = f(x) + f(1) + x - x -1 \ \Rightarrow \ f(1) = 1.$$

and from one point of view

$$(f(\alpha x))' = \alpha f'(\alpha x)$$

from other side,

$$(f(\alpha x))' = f'(x) + \alpha - 1.$$

So we get

$$\alpha f'(\alpha x) = f'(x) + \alpha - 1$$

and

$$f'(\alpha x) = \frac{1}{\alpha}f'(x) + 1 - \frac{1}{\alpha}$$

We can see every $x$ greater than $1$ as $\alpha\cdot 1, \ \alpha>0$:

$$f'(\alpha 1) = \frac{1}{\alpha}f'(1) + 1 - \frac{1}{\alpha}$$

$$f'(\alpha 1) = \frac{4}{\alpha} + 1 - \frac{1}{\alpha} = 1 + \frac{3}{\alpha},$$

which is always bigger than 1, so $f(x)$ grows after $1$.

The only possibility for $f(x_0)<f(1)=1$ is $x_0<1$, i.e. $x_0 \in (0,1)$ according to conditions.

• Please elaborate it! I can't understand it after you calculated f(1). Commented May 16, 2016 at 7:59
• @HarryKarwasra I have replaced $y$ with $\alpha$ just to emphasize that it is come constant. Then, I took the derivative of a composite function $f(g(x))$, where $g(x)=\alpha x$, using chain rule: $\frac{d}{dx}f(g(x)) = f'(g(x))g'(x)$. This yields $f'(\alpha x)\cdot \alpha$. From the other side, we know that $f(\alpha x) = f(x) + f(\alpha)+\alpha x - x - \alpha$, so we can take the derivative of the right side. Commented May 16, 2016 at 8:34
• @HarryKarwasra Thus we obtain the expression for $f'(\alpha x)$ from these two equalities. On the second step we take $x=1$ and remember that $\alpha$ is arbitrary number. So we actually get the solution for derivative of $f(x)$: $$f'(\alpha) = 1 + \frac{3}{\alpha}$$ As long as this expression is always $>0$, it means that $f(x)$ always grows. The value $f(x_0)=0$ cannot be located anywhere on $[1,+\infty)$. Commented May 16, 2016 at 8:41