Why can I not find the top point of this quadratic equation?

**MathIsOhSoHard** · April 29th 2012, 07:27 AM

Given the equation:

$f(x)=1+x-\frac{x^2}{1}-\frac{x^3}{1}$

I want to find the x-value of the top point of this equation, which as a graph I've found to be $x=\frac{1}{3}$

Why can I not find the top point of this quadratic equation?-graphtop.gif

However when I try to find this value, I don't end up with $\frac{1}{3}$

First I differentiate the equation:

$f'(x)=\frac{d}{dx}1+x-\frac{x^2}{1}-\frac{x^3}{1}=1-2\cdot x-3\cdot x^2$

Now I put the differentiated equation equal to 0 and isolate x:

$1-2\cdot x-3\cdot x^2=0$
$x=-1$

Why am I not getting $x=\frac{1}{3}$ ? According to my graph, that is the correct value?

**Quant** · April 29th 2012, 07:38 AM

Hi!

Originally Posted by MathIsOhSoHard

Given the equation:

$f(x)=1+x-\frac{x^2}{1}-\frac{x^3}{1}$

I want to find the x-value of the top point of this equation, which as a graph I've found to be $x=\frac{1}{3}$

Click image for larger version.

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However when I try to find this value, I don't end up with $\frac{1}{3}$

First I differentiate the equation:

$f'(x)=\frac{d}{dx}1+x-\frac{x^2}{1}-\frac{x^3}{1}=1-2\cdot x-3\cdot x^2$

Now I put the differentiated equation equal to 0 and isolate x:

$1-2\cdot x-3\cdot x^2=0$
$x=-1$

How did you do that?

Originally Posted by MathIsOhSoHard

Why am I not getting $x=\frac{1}{3}$ ? According to my graph, that is the correct value?

You can find out by substituting 1/3 for x

$=1-2\cdot (\frac{1}{3})-3\cdot (\frac{1}{3})^2$

$=1- (\frac{2}{3})-3\cdot \frac{1}{9}$

$=1- (\frac{2}{3})-\cdot \frac{3}{9}$

$=1- (\frac{2}{3})-\cdot \frac{1}{3} = 0$

So x=1/3 is a solution of the quadratic equation.

By the way are you sure x=1/3 is the one and only solution?

**MathIsOhSoHard** · April 29th 2012, 07:55 AM

Originally Posted by Quant

By the way are you sure x=1/3 is the one and only solution?

No, there are probably more solutions to it. However I really want to know how to get to 1/3.

**biffboy** · April 29th 2012, 08:08 AM

Your quadratic factorises (1+x)(1-3x)=0 So x=1/3 or -1

**Quant** · April 29th 2012, 08:22 AM

There are many ways to get this solutions. One of them:

You need to solve

$1-2x-3x^2 = 0$

We devide by -3

$-\frac{1}{3}+\frac{2}{3}x+x^2 = 0$

We add a zero

$-\frac{1}{3}+\frac{2}{3}x+(\frac{2}{3 \cdot 2})^2-(\frac{2}{3 \cdot 2})^2+x^2 = 0$

Some rearranging

$-\frac{1}{3}+\frac{2}{3}x+(\frac{2}{6})^2+x^2 = +(\frac{2}{6})^2$

$-\frac{1}{3}+\frac{2}{3}x+(\frac{2}{6})^2+x^2 = \frac{4}{36}$

$\frac{2}{3}x+(\frac{2}{6})^2+x^2 = \frac{4}{36} +\frac{1}{3}$

$\frac{2}{3}x+(\frac{2}{6})^2+x^2 = \frac{4}{36} +\frac{1}{3}$

$\frac{2}{3}x+(\frac{1}{3})^2+x^2 = \frac{4}{36} +\frac{1}{3}$

As you know it is $(a+b)^2 = a^2+2ab+b^2$ (do you know this theoreme? Respectively do you see how we find a and b?)

thus

$\frac{2}{3}x+(\frac{2}{6})^2+x^2$

is the same as

$(x+\frac{1}{3})^2$

Using this in $\frac{2}{3}x+(\frac{1}{3})^2+x^2 = \frac{4}{36} +\frac{1}{3}$

leads us to

$(x+\frac{1}{3})^2 = \frac{4}{36} +\frac{1}{3}$

and so on...

$(x+\frac{1}{3})^2 = \frac{1}{9} +\frac{1}{3}$

$(x+\frac{1}{3})^2 = \frac{1}{9} +\frac{3}{9}$

$(x+\frac{1}{3})^2 = \frac{4}{9}$

$x+\frac{1}{3} = \sqrt{\frac{4}{9}}$

$x+\frac{1}{3} = \pm \frac{2}{3}$

$x = -\frac{1}{3} \pm \frac{2}{3}$

so -1 and 1/3 are the solutions you are looking for

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