# Consider the functional equation f(x-y)=f(x)/f(y) . If f'(0) = p and f'(5)=q , then find f'(-5) ?

May 2, 2017

$f ' \left(- 5\right) = {p}^{2} / q$

#### Explanation:

Making $x = y$ we have

$f \left(0\right) = 1$

Making $x = 2 y$ we have

$f \left(y\right) = f \frac{2 y}{f} \left(y\right)$ or

$f \left(2 y\right) = f {\left(y\right)}^{2}$

or

$f \left(k y\right) = f {\left(y\right)}^{k}$ for $k \in \mathbb{Z}$

but also with $x = 0$

$f \left(- y\right) = f {\left(y\right)}^{-} 1$ and

$f \left(- k y\right) = f {\left(y\right)}^{-} k$

Supposing now

$f \left(y\right) = {a}^{y}$

we have

$\frac{d}{\mathrm{dy}} f \left(y\right) = {a}^{y} {\log}_{e} a \to {a}^{0} {\log}_{e} a = p$

or

$a = {e}^{p}$

and

$f \left(y\right) = {\left({e}^{p}\right)}^{y}$

then

$f ' \left(5\right) = {\left({e}^{p}\right)}^{5} p = q$ and

$f ' \left(- 5\right) = \frac{p}{{e}^{p}} ^ 5$

or

$f ' \left(- 5\right) = {p}^{2} / q$

May 2, 2017

${p}^{2} / q .$

#### Explanation:

Given that, $f \left(x - y\right) = f \frac{x}{f} \left(y\right) \ldots \ldots . \left(\star\right) .$

x=y, &, (star) rArr f(0)=f(y)/f(y)=1.......(1).

x=0, y=y, & (star) rArr f(-y)=f(0)/f(y)=1/f(y).........(2).

x=x, y=-y, & (star) rArr f(x+y)=f(x)/f(-y)=f(x)f(y)

$\therefore , \text{ by } \left(2\right) , f \left(x + y\right) = f \left(x\right) f \left(y\right) \ldots \ldots \ldots \ldots . \left(3\right) .$

Knowing that, $f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x + h\right) - f \left(x\right)}{h} ,$ we have, by $\left(3\right) ,$

$f ' \left(x\right) = {\lim}_{h \to 0} \frac{f \left(x\right) f \left(h\right) - f \left(x\right)}{h} = f \left(x\right) \left\{{\lim}_{h \to 0} \frac{f \left(h\right) - 1}{h}\right\} \ldots \left(\ast\right) .$

Now, f'(0)=p, (ast), & (1) rArr p=lim_(h to 0) (f(h)-1)/h...(4).

Similarly, f'(5)=q, (ast), & (4) rArr q=p*f(5)......(5).

"Finally, by "(ast), &, (4), f'(-5)=p*f(-5)

$= p \cdot \left(\frac{1}{f} \left(5\right)\right) , \ldots \ldots \ldots \ldots \left[\because , \left(2\right)\right]$

$= \frac{p}{\frac{q}{p}} , \ldots \ldots \ldots \ldots \ldots . \left[\because , \left(5\right)\right]$

$\therefore f ' \left(- 5\right) = {p}^{2} / q ,$ as Respected Cesareo R., has derived!

Enjoy Maths.!