Existense of Tangent Line

In summary, the question is whether the graph of f(x) = x^(1/2) has a point of tangency with the line y = (x/4) + 1. To determine this, the limits of the equations at x = 4 must be compared. The curve and tangent line only intersect at (4,2), so this would be the only possible point of tangency. The derivative of both equations at any point on the real axis must also be compared to determine if they have the same gradient.
  • #1
PsychonautQQ
784
10

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...
 
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  • #2
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.
 
  • #3
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?
 
  • #5
A tangent line to a curve at a point has the same gradient as the curve locally.
PsychonautQQ said:
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?
 
  • #6
CAF123 said:
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.
 

1. What is a tangent line?

A tangent line is a straight line that touches a curve at only one point, called the point of tangency. It is perpendicular to the curve at that point, meaning it forms a 90 degree angle with the curve.

2. How is a tangent line different from a secant line?

A secant line is a line that intersects a curve at two or more points, while a tangent line only touches the curve at one point. Additionally, a secant line may not be perpendicular to the curve at the point of intersection, unlike a tangent line.

3. What is the purpose of finding a tangent line?

Finding a tangent line allows us to determine the slope of a curve at a specific point. This can be useful in many applications, such as calculating rates of change or optimizing functions.

4. Can a tangent line exist for all types of curves?

Yes, a tangent line can exist for all types of curves, as long as the curve is smooth and continuous. This includes lines, circles, parabolas, and other more complex curves.

5. How do you find the equation of a tangent line?

The equation of a tangent line can be found using the derivative of the curve at the point of tangency. The derivative gives the slope of the curve at a given point, and this slope can be used in the point-slope formula to find the equation of the tangent line.

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