The minimum of the function

**Patrick_John** · July 2nd 2007, 04:38 PM

Hello

The minimum of the function $f(x) = (log_{2}x)^2 + log_{4}x + 1$ is ?

What's the process to solve this problem?

Thanks.

**Jhevon** · July 2nd 2007, 04:43 PM

Originally Posted by Patrick_John

Hello

The minimum of the function $f(x) = (log_{2}x)^2 + log_{4}x + 1$ is ?

What's the process to solve this problem?

Thanks.

the process is the same as for all problems of this type, find it's derivative and set it equal to zero. if necessary, use the second derivative to verify which of the critical points is a minimum

you will need the change of base formula to change all the logs to ln though

Change of Base Formula for Logarithms:

$\log_a b = \frac { \log_c b}{ \log_c a}$

or maybe noticing that $\log_4 x = \log_2 \sqrt {x}$ will simplify the problem a bit. but i'd go with my first suggestion

**ThePerfectHacker** · July 2nd 2007, 06:24 PM

Originally Posted by Patrick_John

Hello

The minimum of the function $f(x) = (log_{2}x)^2 + log_{4}x + 1$ is ?

What's the process to solve this problem?

Thanks.

Do what Jhevon did,
$\log_2^2 x + \log_2 \sqrt{x} + 1$

Rewrite as,
$\log_2^2 x + \frac{1}{2}\log_2 x+1$

Let $y=\log_2^2 x$ to get,

$y^2 + \frac{1}{2}y+1$

This is a parabola, you can use the formula to obtain its minimum.

**Jhevon** · July 2nd 2007, 06:48 PM

Originally Posted by ThePerfectHacker

Do what Jhevon did,
$\log_2^2 x + \log_2 \sqrt{x} + 1$

Rewrite as,
$\log_2^2 x + \frac{1}{2}\log_2 x+1$

Let $y=\log_2^2 x$ to get,

$y^2 + \frac{1}{2}y+1$

This is a parabola, you can use the formula to obtain its minimum.

Thanks for that TPH, I forgot Patrick isn't in Calculus. I see "minimum" and i switched to calc mode. doing it using calc is fun though, the precalc method was so anticlimactic

**Patrick_John** · July 3rd 2007, 12:49 AM

Thanks for the help.

But since I'm not very aware of what is a minimum, I don't know how to get it using the formula, could somebody explain in more detail how to do it?

Also, why did you rewrite $\log_2 \sqrt{x}$ as $\frac{1}{2}\log_2 x$ ?

**Soroban** · July 3rd 2007, 05:16 AM

Hello, Patrick_John!

I will assume that we are not allowed to use Calculus . . .

Find the minimum of the function: . $f(x) \:= \:(\log_{2}x)^2 + \log_{4}x + 1$

The logs have two different bases; we'll make them the same.

Let $\log_4x = P$

. . Then: . $4^P = x\quad\Rightarrow\quad (2^2)^P = x\quad\Rightarrow\quad 2^{2P} = x$

. . Take logs (base 2): . $\log_2\left(2^{2P}\right) = \log_2x\quad\Rightarrow\quad2P\!\cdot\!\log_22 = \log_2x$

. . Then: . $2P = \log_2x\quad\Rightarrow\quad P = \frac{1}{2}\log_2x$
. . Hence: . $\log_4x = \frac{1}{2}\log_2x$

Substitute into the original equation: . $f(x)\:=\:\left(\log_2x\right)^2 + \frac{1}{2}\log_2x + 1$

Let $z = \log_2x$
Then we have: . $f(z) \:=\:z^2 + \frac{1}{2}z + 1$

This is an up-opening parabola; its minimum is at its vertex.

The vertex formula is: . $\frac{\text{-}b}{2a}$
We have: . $a = 1,\:b = \frac{1}{2}$
Hence, the vertex is at: . $z \:=\:\frac{\text{-}\frac{1}{2}}{2(1)} \:=\:-\frac{1}{4}$

Therefore, the minimum is: . $f\left(\text{-}\frac{1}{4}\right)\;=\;\left(\text{-}\frac{1}{4}\right)^2 + \frac{1}{2}\left(\text{-}\frac{1}{4}\right) + 1 \;=\;\boxed{\frac{15}{16}}$

Thread: The minimum of the function

LinkBack

Thread Tools

Display

The minimum of the function

Similar Math Help Forum Discussions

[SOLVED] Maximum and minimum of function

Minimum value of function

Minimum value for the function:

how to prove a function has a minimum

Minimum Value of Function

Search Tags