Let f:R R be a differentiable function such that f((x+y)/(3))=(f(x)+f(y))/(3) for all x,y …

Question

Let $f:R	o R$ be a differentiable function such that $f(\frac{x+y}{3})=\frac{f(x)+f(y)}{3}$ for all $x,y\in R$, and ${f}^{'}(0)=3$. Then the minimum value of the function $g(x)=3+{e}^{x}f(x)$, is:

Accepted Answer

Given

$f(\frac{x+y}{3})=\frac{f(x)+f(y)}{3}\, x,y\in R,$

and

${f}^{'}(0)=3.$

We have to find the minimum value of

$g(x)=3+{e}^{x}f(x).$

Find the form of $f(x)$: We are given a functional equation. We will use it carefully.

Let

$\frac{x+y}{3}=t.$

Then

$x+y=3t.$

The relation becomes

$f(t)=\frac{f(x)+f(y)}{3}\, x+y=3t.$

We now choose suitable values.

Put $x=y=0$
Then

$f(0)=\frac{f(0)+f(0)}{3}=\frac{2f(0)}{3}.$

So,

$3f(0)=2f(0)\, \implies \, f(0)=0.$

Put $y=0$
Then

$f(\frac{x}{3})=\frac{f(x)+f(0)}{3}=\frac{f(x)}{3}.$

Thus,

$f(x)=3f(\frac{x}{3}).$

Now differentiate both sides with respect to $x$:

${f}^{'}(x)=3\cdot {f}^{'}(\frac{x}{3})\cdot \frac{1}{3}={f}^{'}(\frac{x}{3}).$

So,

${f}^{'}(x)={f}^{'}(\frac{x}{3}).$

Repeating this,

${f}^{'}(x)={f}^{'}(\frac{x}{3})={f}^{'}(\frac{x}{{3}^{2}})=\cdots ={f}^{'}(\frac{x}{{3}^{n}}).$

As $n	o \infty$,

$\frac{x}{{3}^{n}}	o 0.$

Since $f$ is differentiable, ${f}^{'}$ is defined at $0$, so we get

${f}^{'}(x)={f}^{'}(0)=3.$

Hence,

${f}^{'}(x)=3\, x.$

Therefore,

$f(x)=3x+c.$

But $f(0)=0$, so

$c=0.$

Thus,

$f(x)=3x.$

Write $g(x)$: Now

$g(x)=3+{e}^{x}f(x)=3+{e}^{x}(3x)=3+3x{e}^{x}.$

So,

$g(x)=3(1+x{e}^{x}).$

Minimise $x{e}^{x}$: Let

$h(x)=x{e}^{x}.$

Then

${h}^{'}(x)={e}^{x}+x{e}^{x}={e}^{x}(1+x).$

Since ${e}^{x}>0$ for all $x$, we have

${h}^{'}(x)=0\, \iff \, 1+x=0\, \iff \, x=-1.$

Now,

for $x
for $x>-1$, ${h}\^{'}(x)>0$, so $h$ is increasing.

Hence $h(x)$ has a minimum at $x=-1$.

Its minimum value is

$h(-1)=(-1){e}\^{-1}=-\frac{1}{e}.$

Therefore,

${g}\_{min}=3(1-\frac{1}{e})=3(\frac{e-1}{e}).$

Final Answer
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Solutions · 1