1. ## Tangents and Normals

Hi i need to work out the equation of the tangent and norm of the following function.
-(25-x^2)^1/2 where x=4

I'm pretty sure i have messed up but here are my solutions that i got.
y= +/- 3
F'(x)= (-4*9^1/2)/9
norm being 9/(4*9^1/2)

If i am correct up to here my equations should be right. Help is greatly appreciated

2. Hello,

Originally Posted by Webby
Hi i need to work out the equation of the tangent and norm of the following function.
-(25-x^2)^1/2 where x=4

I'm pretty sure i have messed up but here are my solutions that i got.
y= +/- 3
F'(x)= (-4*9^1/2)/9
norm being 9/(4*9^1/2)

If i am correct up to here my equations should be right. Help is greatly appreciated
When differentiating a function, you can't just put the values... especially here, where it's a nonsense to put the values of y in f'(x)

The equation of a tangent where x=a is : $\displaystyle y=f'(a)(x-a)+f(a)$

So calculate f'(x) :
$\displaystyle f(x)=-(25-x^2)^{1/2}$

$\displaystyle f'(x)=x(25-x^2)^{-1/2}=\frac{x}{\sqrt{25-x^2}}$ (I'll let you do the calculation... just tell if you can't).

--> $\displaystyle f'(4)=\frac{4}{\sqrt{25-4^2}}=\frac 43$

$\displaystyle f(4)=-(25-4^2)^{1/2}=-\sqrt{25-4^2}=-3$, and not $\displaystyle \pm 3$, because a square root is always positive. So -sqrt(25-4²)<0

3. ## Even I am confused

Originally Posted by Moo
because a square root is always positive. So -sqrt(25-4²)<0
[x^2]^1/2 = x if x>0 and -x if x<0 e.g [(-3)^2]^1/2=-3 but
(-3)^2=9 so if we put this talue in previous eq we get
9^1/2=-3 so we get square root may give a negative value.
2]I think square root of a number means to find a number say a which when multiplied to itself gives the same number e.g 4^1/2 means to find a number say a which when multiplied to itself gives 4 so a)2x2=4 and b)-2x-2=4 so 4^1/2 may also give -2 which is not positive.
Though -(x)^1/2 is <0 but it does not mean square root is always positive.if i am wrong please point my mistake out. Thanks

4. I think there is something wrong...

Yes, $\displaystyle x^2=9 \implies x=\pm 3$, but it doesn't mean that $\displaystyle \sqrt{9}=\pm 3$.

(Of course, I'm not talking about the imaginary numbers and all this stuff)

This is the curve of $\displaystyle \sqrt{x}$

Originally Posted by Wikipedia
If otherwise unqualified, "the square root" of a number refers to the principal square root
Originally Posted by Wikipedia
Every non-negative real number x has a unique non-negative square root, called the principal square root and denoted with a radical symbol as √x.
----------------------

Originally Posted by nikhil
[x^2]^1/2 = x if x>0 and -x if x<0 e.g [(-3)^2]^1/2=-3 but
(-3)^2=9 so if we put this talue in previous eq we get
9^1/2=-3 so we get square root may give a negative value.
2]I think square root of a number means to find a number say a which when multiplied to itself gives the same number e.g 4^1/2 means to find a number say a which when multiplied to itself gives 4 so a)2x2=4 and b)-2x-2=4 so 4^1/2 may also give -2 which is not positive.
Though -(x)^1/2 is <0 but it does not mean square root is always positive.if i am wrong please point my mistake out. Thanks

5. ## Still confused

You have drawn the function
y=x^1/2 but if you draw the graph of relation y=x^1/2 then for each value x two values will be obtained one positive and other negative.
its true that square root of every non negative number is positive. Thats why i created the example [(-3)^2]^1/2=-3
but (-3)^2=9 so if we substitute this value we get
9^1/2=-3 (which is correct)
if I just ask 9^1/2=? Without knowing whether this 9 represent 3^2 or (-3)^2 saying the answer is +3 will be wrong.so if we get x^1/2
in a question as a subpart I don't think it should only be treated as a function,it may also be used as relation.so if You take x^1/2 as relation you will get two values which is against saying that square root is always positive.what do you think??

6. If I draw x^(1/2), it just yield the same graph

$\displaystyle \sqrt{x}=x^{1/2}$ is not define (over reals) for negative numbers.

When you write $\displaystyle ((-3)^2)^{1/2}$, I guess you do first (-3)^2, which gives 9, then, power it to 1/2. And this yields a positive number : 3.

7. ## not enough

Originally Posted by Moo
If I draw x^(1/2), it just yield the same graph

$\displaystyle \sqrt{x}=x^{1/2}$ is not define (over reals) for negative numbers.

When you write $\displaystyle ((-3)^2)^{1/2}$, I guess you do first (-3)^2, which gives 9, then, power it to 1/2. And this yields a positive number : 3.
Actually when I said draw the graph of y=x^1/2 I did not meant to involve negative value (as we are talking about real values in real plane not about imaginary quantities in argand plane) I meant you will get graph similar to y^2=4ax where a>0 (for just a rough visual begaviour) that is for x>0 you will get two values like 4^1/2
=2 and -2 which is impossible for a function(as a function can not have two values for same element x) but this is not a problem if x^1/2 is treated as relation because getting two values for same element x will be permissible so again 9^1/2 may also mean -3.
Dude like your guess that (-3)^2 gives 9 and 9^1/2 gives +3
BUT [(-3)^2]^1/2 is actually =-3 but not=3 so the precedure you gave is not correct.
Again think if -2x-2=4 then why 4^1/2 can not yield -2 which is a negative quantity against the saying square root is always positive.

8. If a problem is *given* to you as $\displaystyle f(x)=\sqrt{x}$ then only nonnegative answers are in the range...that's the definition of $\displaystyle \sqrt{x}$ as Moo pointed out.

Of course

$\displaystyle 4 = x^2$

for example, there is no such restriction...$\displaystyle x$ can be positive or negative, that's why when you "take the root" of both sides, you must account for both possibilities:

$\displaystyle 2 = |\sqrt{x^2}|$

$\displaystyle 2 = |x|$

$\displaystyle \pm2 = x$

but notice that $\displaystyle \sqrt{x^2}$ is still defined as nonnegative.

So this should even apply to subproblems...as long as you correctly account for +/- when you take the root, $\displaystyle \sqrt{x}$ will always be positive.

Eric

9. ## Welcome Eric

Hi! Eric, its very nice to have you in the forum.
So do you mean [(-3)^2]^1/2 is =3 that is if we square a negative number and that square root it, the number will become positive?

10. Originally Posted by nikhil
Hi! Eric, its very nice to have you in the forum.
So do you mean [(-3)^2]^1/2 is =3 that is if we square a negative number and that square root it, the number will become positive?
Yes $\displaystyle \sqrt{x^2}\equiv|x|$

If $\displaystyle x<0$ then $\displaystyle |x|=-x$

and since x is negative $\displaystyle -x>0$

11. Originally Posted by Mathstud28
Yes $\displaystyle \sqrt{x^2}\equiv|x|$
Hey! That will definetly help alot. But if -2x-2=4
Then why can't 4^1/2 can not be taken as -2 .
NOTE:considering y=x^1/2
as a function is not necessary,poster may take it as a relation also]

12. Originally Posted by nikhil
Hey! That will definetly help alot. But if -2x-2=4
Then why can't 4^1/2 can not be taken as -2 .
NOTE:considering y=x^1/2
as a function is not necessary,poster may take it as a relation also]
Because it is a number, not a variable, it is convention to do as such.

Someone else might have a more enlightening response.

13. Originally Posted by nikhil
Hey! That will definetly help alot. But if -2x-2=4
Then why can't 4^1/2 can not be taken as -2 .
NOTE:considering y=x^1/2
as a function is not necessary,poster may take it as a relation also]
We usually use the notation $\displaystyle x^{m/n}$ to mean $\displaystyle \sqrt[n]{x^m}$, i.e., the principal $\displaystyle n^{\rm th}$ root.

Thus, $\displaystyle 4^{1/2} = \sqrt4 = 2$. But if you were solving the equation $\displaystyle x^2 = 4$ you would have to take into account the possibility of $\displaystyle x$ being negative, so the solution is $\displaystyle x = \pm\sqrt4 = \pm2$.