maximum value of func.

**stud_02** · July 12th 2009, 08:00 AM

$f(x)=2(a-x)(\sqrt{x^2+b^2} + x)$

for all real x,
show the maximum value of f(x) is $( a^2+b^2)$
and $x=\frac{a^2-b^2}{2a}$ when f(x) is maximum

**Rapha** · July 12th 2009, 06:41 PM

Hi stud_02!

Originally Posted by stud_02

$f(x)=2(a-x)(\sqrt{x^2+b^2} + x)$

for all real x,
show the maximum value of f(x) is $( a^2+b^2)$
and $x=\frac{a^2-b^2}{2a}$ when f(x) is maximum

For all real x? Your Solution seems to be wrong,
if a=0 it does not make any sense

And by the way, if b = 1 and a = 0
$f(x) = -2x(\sqrt{x^2+1}+x)$ has its maximum in $\approx (-0.34 ; -1.66)$

**simplependulum** · July 12th 2009, 11:02 PM

Originally Posted by stud_02

$f(x)=2(a-x)(\sqrt{x^2+b^2} + x)$

for all real x,
show the maximum value of f(x) is $( a^2+b^2)$
and $x=\frac{a^2-b^2}{2a}$ when f(x) is maximum

Substitute
$t = \sqrt{x^2+b^2} + x$
$\implies$
$t^2 - 2tx + x^2 = x^2 + b^2$
$\implies$

$x = \frac{t^2-b^2}{2t}$

Go back to the function ,

we will have
$f(t) = -t^2 + 2at + b^2$

$= -(t-a)^2 + b^2 + a^2$

so it has max. value $a^2 + b^2$ when $t = a$
$x = \frac{a^2-b^2}{2a}$

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