Sum of distances the same as the former

[leprofece]

find a point on the axis OX whose sum of distances to landmarks: (2, 0) and (0, 3) is minimal.

Answer (2,0)

As the title says it is the same as the former
So the equations must be
sqrt((x-2)²+0²)+sqrt((x-0)²+ 9) because the point is (x.0)

but it seems I am wrong because i don't get the answer

 

[MarkFL]

You should get the given answer. But, since you did not post your work, I can't tell you where you may have gone wrong. It is worked very similarly to the last question you posted.

 

[leprofece]

You should get the given answer. But, since you did not post your work, I can't tell you where you may have gone wrong. It is worked very similarly to the last question you posted.

THIS IS MY WORK
So the equations must be
sqrt((x-2)²+0²)+sqrt((x-0)²+ 9) because the point is (x.0)

After deriving i got 1 + x/(sqrt(x²+9)
And I got No answer

but it seems I am wrong because i don't get the answer

 

[I like Serena]

So the equations must be
sqrt((x-2)²+0²)+sqrt((x-0)²+ 9) because the pòint is (x.0)

After deriving i got 1 + x/(sqrt(x²+9)
And I got No answer

but it seems I am wrong because i don't get the answer

That is not quite the right derivative.
The derivative of $\sqrt{(x-2)^2} = |x-2|$ is not just $1$.
Its derivative depends on the sign of $(x-2)$.
Consider the function:
$$g(x)=\sqrt{x^2} = |x|$$
Its derivative is:
$$g'(x)=\begin{cases}
-1 &\text{ if } x < 0 \\
\text{Undefined} &\text{ if } x = 0 \\
+1 &\text{ if } x > 0 \end{cases}$$

Where does $g(x)$ have its minimum?

 

[leprofece]

En -1?

 

[I like Serena]

En -1?

Huh?

Do you mean $x=-1$?

If so, then $|x|=|-1|=1$, which is not the minimum value for $|x|$...

 

[leprofece]

Huh?

Do you mean $x=-1$?

If so, then $|x|=|-1|=1$, which is not the minimum value for $|x|$...

ZERO then i did not take into account the absolute value (Heidy)(Angry)

And Abs( 0-2) is 2
so (Priidu) this is the answer

 

[I like Serena]

ZERO then i did not take into account the absolute value (Heidy)(Angry)

Good!

Note that the derivative of g(x) is undefined at 0.
What actually happens is that the derivative flips from a negative value to a positive value.
At the point where it flips, we have the minimum.

The same applies to your problem.

 

[leprofece]

I realize that the second Function wasnt taken into account so That is sqrt (x² +9)
WHY?
I Think because x / sqrt( ,,,) = 0 it is equal to 0
Am i right??

So the equations must be
sqrt((x-2)2+02)+sqrt((x-0)2+ 9) because the pòint is (x.0)

After deriving i got 1 + x/(sqrt(x2+9)
And I got No answer

but it seems I am wrong because i don't get the answer

 

[I like Serena]

After deriving i got 1 + x/(sqrt(x2+9)
And I got No answer

but it seems I am wrong because i don't get the answer

This is not the correct derivative.
It is only correct for $x>2$.
And then the derivative is always positive.

 

[leprofece]

But Why isnot correct
x/sqrt(...) is the deivative of the other
maybe because if I equate to 0 I get 0?

 

[I like Serena]

But Why isnot correct
x/sqrt(...) is the deivative of the other
maybe because if I equate to 0 I get 0?

The derivative of $\sqrt{(x-2)^2}$ is not $1$.