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

Given the equation:

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

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

Attachment 23730

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

First I differentiate the equation:

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

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

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

$\displaystyle x=-1$

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

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

Hi!

Quote:

Originally Posted by

**MathIsOhSoHard** Given the equation:

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

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

Attachment 23730
However when I try to find this value, I don't end up with $\displaystyle \frac{1}{3}$

First I differentiate the equation:

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

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

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

$\displaystyle x=-1$

How did you do that?

Quote:

Originally Posted by

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

You can find out by substituting 1/3 for x

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

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

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

$\displaystyle =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?

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

Quote:

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.

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

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

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

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

You need to solve

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

We devide by -3

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

We add a zero

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

Some rearranging

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

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

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

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

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

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

thus

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

is the same as

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

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

leads us to

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

and so on...

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

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

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

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

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

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

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