Derivation

[learningmathh]

I have this function f(x)=2Pix^2 + 400/x. I've to derivate it and find the lowest point for x. I've managed this far 4Pix -400/x^2 but how to finish it? Anyone?

 

[HallsofIvy]

The "lowest point for x"? Do you mean the value of x that make f(x) least? I would think that the reason you differentiated it was because you knew that the least value of a function, over all x, occurs where the derivative is 0. Of course, you will also have to check that it really is a minimum rather than a local maximum or inflection point.

 

[Deleted member 4993]

I have this function f(x)=2Pix^2 + 400/x. I've to derivate it and find the lowest point for x. I've managed this far 4Pix -400/x^2 but how to finish it? Anyone?

Set that equal to 0 - and solve for 'x' . You will get 3 values of 'x' - choose the value where 'x' is real and f"(x)≥0 for minimum value.

 

[learningmathh]

The "lowest point for x"? Do you mean the value of x that make f(x) least? I would think that the reason you differentiated it was because you knew that the least value of a function, over all x, occurs where the derivative is 0. Of course, you will also have to check that it really is a minimum rather than a local maximum or inflection point.

Exactly yes

 

[learningmathh]

Set that equal to 0 - and solve for 'x' . You will get 3 values of 'x' - choose the value where 'x' is real and f"(x)≥0 for minimum value.

Well, how to do that if you havent learned how to solve x^3. It must be another way to solve it right? By the way, x is 1<x<7.

 

[Deleted member 4993]

Well, how to do that if you havent learned how to solve x^3. It must be another way to solve it right? By the way, x is 1<x<7.

Are you saying that in your algebra class you have not been taught the following formula:

a³ - b³ = (a - b)(a² + ab + b²)

If the answer is 'yes' - then you ought to sue your secondary school district for utter incompetence!!

By the way, since you have a given domain for the function, you need to check the property of the function at the end points also.

 

[learningmathh]

Are you saying that in your algebra class you have not been taught the following formula:

a³ - b³ = (a - b)(a² + ab + b²)

If the answer is 'yes' - then you ought to sue your secondary school district for utter incompetence!!

By the way, since you have a given domain for the function, you need to check the property of the function at the end points also.

What do you mean? I know that formula bu how can you use it in this function? I've solved for =0 as you said and I got this 4pi x^3 - 400=0, so how can I use that formula?

(by the way how do you Write With symbols?)

 

[Deleted member 4993]

What do you mean? I know that formula bu how can you use it in this function? I've solved for =0 as you said and I got this 4pi x^3 - 400=0, so how can I use that formula?

(by the way how do you Write With symbols?)

$\displaystyle 4\pi x^3 - 400 = 0$

$\displaystyle x^3 -\ \left [\sqrt[3]{\dfrac{100}{\pi}}\right ]^3 = 0$

You really ought to sue that school district!!

 

[learningmathh]

$\displaystyle 4\pi x^3 - 400 = 0$

$\displaystyle x^3 -\ \left [\sqrt[3]{\dfrac{100}{\pi}}\right ]^3 = 0$

You really ought to sue that school district!!

Oh thanks, but are you really serious about that or are you joking??

 

[daon2]

[/B]
Oh thanks, but are you really serious about that or are you joking??

algebra is no joke. better call saul

 

[HallsofIvy]

While solving cubics in general can be very hard, this one,
$\displaystyle 4\pi x^3- 400= 0$, should be very simple, because it does not have the "$\displaystyle x^2$" or "$\displaystyle x$" terms.

First add 400 to both sides: $\displaystyle 4\pi x^3= 400$.

Now divide both sides by $\displaystyle 4\pi$: $\displaystyle x^3= \frac{100}{\pi}$.

Finally take the cube root of both sides: $\displaystyle x= \sqrt[3]{\frac{100}{\pi}}$.

Every real number has 3 cube roots- one a real number, the other two complex conjugates.

 

[learningmathh]

While solving cubics in general can be very hard, this one,
$\displaystyle 4\pi x^3- 400= 0$, should be very simple, because it does not have the "$\displaystyle x^2$" or "$\displaystyle x$" terms.

First add 400 to both sides: $\displaystyle 4\pi x^3= 400$.

Now divide both sides by $\displaystyle 4\pi$: $\displaystyle x^3= \frac{100}{\pi}$.

Finally take the cube root of both sides: $\displaystyle x= \sqrt[3]{\frac{100}{\pi}}$.

Every real number has 3 cube roots- one a real number, the other two complex conjugates.

Thanks very much, I understood it! I was just blind after doing math several ours straight