Is There a Point of Tangency Between f(x) = x^(1/2) and y = (x/4) + 1?

[PsychonautQQ]

Homework Statement

Q: Does the graph of f(x) = x^(1/2) have a point of tangency with the line y = (x/4) + 1?

Homework Equations

lim x->a (f(x) - f(a)) / (x-a)

The Attempt at a Solution

If the limit exists of the relevant equation than there is a point of tangency.

So I'm having a bit of trouble proving anything here...
using the equation I come to

lim x->a (x^(1/2) - (x/4) + 1) / (x-a)

which looks like as x goes to a the denominator will approach zero which leads me to believe it diverges to infinity? does that mean the limit doesn't exist?

When setting the equations equal to each other I find that they intersect at (4,2). Do I plug 4 in for a? same thing happens except this time the numerator obviously approaches zero as well...

 

[CAF123]

What does the curve y=√x and a line of tangency have in common? You do not have to resort to the precise definition of a limit to solve the question.

 

[PsychonautQQ]

How would I use the precise definition here? Just for the sake of using it. To answer your question the curve y=x^(1/2) and a tangent line have in common? I'm not sure... This equation will cover the whole positive y positive x-axis as x-> infinity but will get more and more linear looking at big values of x? And tangent lines are linear?

 

[ehild]

See http://mathworld.wolfram.com/TangentLine.html

ehild

 

[CAF123]

A tangent line to a curve at a point has the same gradient as the curve locally.

How would I use the precise definition here? Just for the sake of using it.

Try writing the above in math. You are right that the line and the curve intersect only at x=4, so that would be the only possible point of tangency. You want the limit as x → 4 of f(x) = √x to be the same as the limit as x→4 of g(x) = (x/4) + 1.

To answer your question the curve y=x^(1/2) and a tangent line have in common? I'm not sure...

See above.

You could do the same as the above, but without invoking the limit. What is the derivative (i.e gradient) of y=√x at any point along the real axis? What about y=(x/4)+1?

 

[CAF123]

You want the limit as x → 4 of f(x) = √x to be the same as the limit as x→4 of g(x) = (x/4) + 1.

This should read: you want the limit as x→4 of (f(x)-f(4))/(x-4) and the limit as x→4 of (g(x)-g(4))/(x-4), where f and g are defined in the quote.