Dumb slope question

1. Aug 20, 2003

mmapcpro

Hi,

I was looking at a problem from an old test and I got confused about something.

Q. Find the slope of the tangent line to the given curve at a given point:

y = x(x^2 + 3)^1/2 ; P(1,2)

S. First I found the derivative:
[(x^2 + 3)^1/2] + [(x^2)(x^2 + 3)^-1/2]

Plugged in the values for x and y at the given point:
[(4)^1/2] + [(4)^-1/2]

a) 2 b) -5/2 c) 3/2 d) 5/2 e) none of these

Square root of 4 can be + OR - 2, no?

That would mean I can have 4 possible solutions:
(4/2) + (1/2) = 5/2
(4/2) + (-1/2) = 3/2
(-4/2) + (1/2) = -3/2
(-4/2) + (-1/2) = -5/2

Since 3 of these are listed as choices, how did I pick the right one? (I chose d)5/2 and it was marked correct)

Forgive me if this is shockingly stupid...I am trying to review all my calculus and physics to start classes up again in the winter.

2. Aug 20, 2003

mmapcpro

NEVERMIND!!!!

I feel dumb...

Of course, the original equation must also hold true, which means that (4)^1/2 can only be +2

3. Aug 21, 2003

HallsofIvy

Staff Emeritus
In addition, "square root" is a FUNCTION which means it can have only one value. The square root of any positive number, a, is, by definition, the POSITIVE number, x, such that x2= a.

Of course, x2= a has TWO solutions: they are [sqrt](a) and -[sqrt](a).

4. Aug 24, 2003

When you say FUNCTION, do you actually mean CONTINOUS ONE-TO-ONE FUNCTION?

5. Aug 24, 2003

HallsofIvy

Staff Emeritus
When I say FUNCTION, believe it or not, I mean FUNCTION. I do not meant "continuous" since that is not part of the definition of FUNCTION, although, in this problem, since the original post assumed the function was differentiable, it must be continuous. Nor do I mean "one-to-one". I consider "f(x)= x<sup>2</sup>", which is NOT one-to-one (f(2)= f(-2)), a perfectly good function. I do require that functions on the real numbers be "well-defined"- that is, that a single value of x can give only one value of f(x). That is, after all, the distinction between a "function" and "relation".