Find Slope of Tangent Line at P(1,2): Answer = 5/2

[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]

The choices for answers were:
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.

 

[mmapcpro]

NEVERMIND!

I feel dumb...

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

 

[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).

 

[On Radioactive Waves]

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]

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)= x²", 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".