# Dumb slope question

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.

NEVERMIND!!!!

I feel dumb...

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

HallsofIvy
Homework Helper
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).

Originally posted by HallsofIvy
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).

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

HallsofIvy