Dedicated Mathematician

[EvoN1020v]

Anybody please?

[Klaynos]

Well when the differntial of a function = 0 then it is a max or a min, so take

y=x²

 

This differntiated becomes:

 

2x

 

This is equal to 0 when x = 0 So the point is at x=0 If you then place this back into the original eqation we see the coorinates of this point are (0,0)

 

To find whether this is a max or min we differntiate the function a second time makeing:

 

2

 

If this is >0 then it is a minima, <0 Then a maxima, and =0 it is a point of inflection...

[EvoN1020v]

I know that the devirative of [math]y=x^2[/math] is [math]2x[/math]. I am looking for something more complicate than that. How about [math]y=x^2-2x-15[/math]?

 

The factioned x values are [math](x-5)(x+3)[/math], so the value of x's are 5 and -3.

 

So the minimum of the quadratic equation is [math]-17[/math]. Am I correct? Hmmm.. let me check my graphic calculator. Nope, I'm wrong. The answer is actually -16. Where did I go wrong? Should I be putting the "1" in the quadratic equation or the factioned values?

 

Because the factioned values did work. [math](x-5)(x+3)[/math] is: [math](1-5)(1+3)[/math], therefore [math]-16[/math].

 

The question here is: What is the devirative for the quadratic equation, and explain to me from that, how to find the actual value.

 

[Connor]

you would use the same process:

 

[math]y=x^2-2x-15[/math]

 

[math]y'=2x-2[/math]

 

[math]0=2x-2[/math]

 

[math]x=1[/math]

 

so there's a extreme at x=1

 

[math]y''=2>0[/math]

 

so it is a minimum

 

plug that back into the original equation and there's a minimum at (1,-16)

[m4rc]

The derivative of [math]x^2-2x-15[/math] is [math]2x-2[/math] (because the derivative of a sum of functions is equal to the sum of the derivatives of each function).

 

The slope of any function will be zero at its maximum/minimums, therefore the derivative of the function will be zero.

So [math]2x-2=0[/math] leading to [math]x=1[/math]

 

We can conclude that [math]x^2-2x-15[/math] is horizontal at [math]x=1[/math]. This still does not indicate it is a maximum, minimum or neither.

 

To be certain you should calculate the second derivative of y.

The second derivative of [math]x^2-2x-15[/math] is the derivative of [math]2x-2 [/math]which is 2. If the second derivative is positive then you have a minimum, if the second derivative is negative then you have a maximum.

[EvoN1020v]

Ah, thanks Connor and m4rc.

 

Let me restate your explanations:

 

Let [math]f(x)=(x-5)[/math] and [math]g(x)=(x+3)[/math].

 

[math]h(x)=f(x)+g(x)[/math]

 

[math](x-5)+(x+3)[/math]

 

[math]2x-2[/math]

 

[math]x=1[/math].

 

So the minimum would be [math](1, -16)[/math].

 

It's easy to determine if it's a minimum or maximum because you look at the sign of the [math]x^2[/math]. If it's a positive, then you're looking at a minimum, and if it's negative, then you are looking at a maximum.

 

 

What do you mean that the second derivative of [math]2x-2[/math] is 2?

[m4rc]

no, you calculated the derivative wrong.

 

if [Math]y(x)=x^2-2x-15[/Math]

[Math]y(x)=f(x)+g(x)+h(x)[/Math] where [Math]f(x)=x^2[/Math], [Math]g(x)=-2x[/Math] and [Math]h(x)=-15[/Math]

 

 

[Math]dy/dx=df/dx+dg/dx+dh/dx[/Math]

[Math]=2x-2+0[/Math]

[m4rc]

Ah' date=' thanks [b']Connor [/b]and m4rc.

What do you mean that the second derivative of [math]2x-2[/math] is 2?

 

The second derivative of [Math]x^2-2x-15[/Math] is 2,

means that the derivative of the derivative of [Math]x^2-2x-15[/Math] is 2.

[EvoN1020v]

no' date=' you calculated the derivative wrong.

 

if [Math']y(x)=x^2-2x-15[/Math]

[Math]y(x)=f(x)+g(x)+h(x)[/Math] where [Math]f(x)=x^2[/Math], [Math]g(x)=-2x[/Math] and [Math]h(x)=-15[/Math]

 

 

[Math]dy/dx=df/dx+dg/dx+dh/dx[/Math]

[Math]=2x-2+0[/Math]

 

Would you say that [math]\frac{dy}{dx}[/math] is [math]\frac{dy}{dx} = \frac{x^2}{x} + \frac{-2x}{x} + \frac{-15}{x}[/math]?

 

The derivative of [math]x^2[/math] is [math]2x[/math]. Then, cross the x out of the [math]\frac{-2x}{x}[/math] yields [math]-2[/math]. Then the last one is an infinite series, so it produces a [math]0[/math]. Therefore, [math]2x-2+0[/math].

 

Am I correct?

[Klaynos]

It's easy to determine if it's a minimum or maximum because you look at the sign of the [math]x^2[/math]. If it's a positive' date=' then you're looking at a minimum, and if it's negative, then you are looking at a maximum.

[/quote']

 

For these simple functions this is true but for more advanced ones it can become annoying and the second dirivative is a usefull shortcut/check...

[EvoN1020v]

You didn't answer my question in post #109? Am I correct by my calculation provided or not?

[Connor]

no, you can't do that, and it just makes things harder. It's easier just to calculate the derivative straight out

[m4rc]

Would you say that [math]\frac{dy}{dx}[/math] is [math]\frac{dy}{dx} = \frac{x^2}{x} + \frac{-2x}{x} + \frac{-15}{x}[/math]?

 

The derivative of [math]x^2[/math] is [math]2x[/math]. Then' date=' cross the x out of the [math']\frac{-2x}{x}[/math] yields [math]-2[/math]. Then the last one is an infinite series, so it produces a [math]0[/math]. Therefore, [math]2x-2+0[/math].

 

Am I correct?

 

No. You seem to be assuming that if [math]y=x^n[/math] then [math]dy/dx=x^n/x [/math] which is not true.

If [math]y=x^n[/math] then dy/dx=nx^(n-1) .

 

For the derivative of many functions see http://mathworld.wolfram.com/Derivative.html

[EvoN1020v]

Don't you mind to show me the solution? How it works?

[m4rc]

Don't you mind to show me the solution? How it works?

 

If you mean that you would like to see the proof for the rule for determining the derivative of a polynomials then see http://en.wikipedia.org/wiki/Calculus_with_polynomials .