Anybody please?

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...

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.

 

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)

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.

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?

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]

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.

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?

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...

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

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

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

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

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 .