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I am trying to find the tangent line at $y=\sqrt{x} $ , (1,1)

I know that I need to use the tangent line equation and I end up with $(\sqrt{x} - 1)/1-1$

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    Are you trying to find the tangent line or the _slope_ of the tangent line? There is a big difference. The tangent line has an equation ($y = mx + b$ or something), and the _slope_ of the tangent line is a number.2011-09-11
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    Note that the equation of a line will have shape $ax+by=c$, where $a$, $b$, $c$ are constants. Anything that *cannot be put* into that shape cannot be right.2011-09-11
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    "Find an equation of the tangent line to the curve at the given point." Not sure what the differences are in what you mentioned.2011-09-11
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    @Jordan: I believe André is saying that there is a difference between "an _equation_ of the tangent line to a curve" and the _number_ given by $\lim_{x\to 1}\frac{\sqrt{x} - 1}{x - 1}.$2011-09-11

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EDIT: Let's clarify a couple of things.

The slope of the secant line between $(a, f(a))$ and $(x,f(x)))$ is $$\frac{f(x) - f(a)}{x-a}.$$

The slope of the tangent line at $(a, f(a))$ is $$\lim_{x\to a}\frac{f(x) - f(a)}{x-a}.$$

To find the equation of a tangent line, one needs to use the point-slope formula, which I've explained below.


Now, in your case, $f(x) = \sqrt{x}$, and we have $a = 1$, $f(a) = 1$. So the slope of the tangent line is $$\lim_{x\to 1}\frac{\sqrt{x} - 1}{x-1}.$$ Now we have to evaluate this limit.

If we try to evaluate this limit by just plugging in $x = 1$, we get $0/0$, which is a problem (dividing by zero is bad), so we need a new strategy.

Idea: When evaluating the limits of fractions, a good trick is to multiply the top and bottom by the "radical conjugate." So:

$$\begin{align} \frac{\sqrt{x} - 1}{x-1} & = \frac{\sqrt{x} - 1}{x-1}\frac{\sqrt{x} + 1}{\sqrt{x} + 1} \\ & = \frac{(\sqrt{x} - 1)(\sqrt{x} + 1)}{(x-1)(\sqrt{x} + 1)} \\ & = \frac{x - 1}{(x-1)(\sqrt{x} + 1)} \\ & = \frac{1}{\sqrt{x} + 1}. \end{align}$$

Now we can evaluate $$\lim_{x \to 1} \frac{\sqrt{x} - 1}{x-1} = \lim_{x\to 1}\frac{1}{\sqrt{x} + 1}$$ by plugging in $x = 1$ no problem. This will give us the slope of the tangent line. If you want the equation of the tangent line, you need the point-slope formula, explained below.


The point-slope formula says that a line with slope $m$ that passes through $(x_0, y_0)$ has an equation of the form $$y - y_0 = m(x-x_0).$$

In your case, the tangent line passes through $(1,1)$, so you can plug in $x_0 = 1$, $y_0 = 1$. We'll also have the slope, $m$, from the previous section once we evaluate that limit (which I leave to you to do).

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    Well in my book is explains to me to by using the equation (f(x)-f(a))/(x-a) I tried to do that and I get something that doesn't make sense.2011-09-11
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    @Jordan: Note that basically $y = f(x)$ and $y_0 = f(x_0)$, so Jesse's answer says $m = (f(x) - f(x_0))/(x - x_0)$. Replacing $x_0$ with $a$ gives the same expression as your book.2011-09-11
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    I see. But are you sure that the formula $(f(x) - f(a))/(x-a)$ really gives you a formula for the tangent line? Because really, $(f(x) - f(a))/(x-a)$ gives the _slope_ of the _secant_ line between $(a, f(a))$ and $(x,f(x))$. On the other hand, the _slope_ of the _tangent_ line is $$\lim_{x\to a} \frac{f(x) - f(a)}{x-a}.$$ Note that neither of these formulas give an equation for a line, but rather a formula for computing the _slope_ of a line.2011-09-11
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    Yes it does, does that mean I make a = x?2011-09-11
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    No, it means that you take the limit as $x \to a$. Perhaps you should review how to take limits.2011-09-11
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    I know limits well enough I am just not sure what the book wants me to do since there is so little explanation.2011-09-11
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    I still don't understand why x does not equal a if the point is (1,1) x = 1 and y = 1 so why is 1 not used in place of x.2011-09-11
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    @Jordan I mean no disrespect, but your suggestion earlier of setting $a = x$ suggests you don't fully understand the concept of a limit.2011-09-11
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    I understand limits, I just don't always understand how they get manipulated to work into these problems. That is why I am doing this homework, so I can learn how they work in these problems. Limits are pretty simple, I just don't know how they work in these problems yet because I haven't done the homework yet.2011-09-11
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    Neither Jesse's nor Austin's comments were rude or offensive. Do not flag them as such.2011-09-11
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    Well it isn't productive to tell me that I am an idiot and then reccomend I waste my time studying something that will just make me fail my test.2011-09-11
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    @Jordan No one has suggested you are an idiot. I think when you say you understand limits, you mean you can solve problems like "Evaluate $\lim_{x \rightarrow a} f(x)$". What I am saying (and I suspect Jesse is saying) is that you don't appear to have a strong grasp of what limits *are*. If your test is in the near future, then perhaps it is best just to cram what you can in a short amount of time. When there is less time pressure, I recommend reading an early section on limits and focusing more on what limits *do* as opposed to how they are *calculated*.2011-09-11
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    I understand what limits do and how they are used, I sometimes just have trouble with using them in these more real world situations involving memorizing sets of equations.2011-09-11
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    @Jordan: Please understand me: when I suggested that you "review how to take limits," I meant that you should review how to calculate them, since you seem to be having trouble with this. Neither Austin nor I are insulting your intelligence; we are both trying to help.2011-09-11
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    @Jordan: If the point in question is $(1,1)$, then this tells us that $a = 1$ and $f(a) = 1$. We don't need to substitute anything in for $x$ or $\sqrt{x}$ since otherwise we could not take the limit as $x \to 1$. And again: after we take the limit, we should get a _number_, not a function nor an equation. To get the _equation_ of the line, we need to use the point-slope formula.2011-09-11
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Several comments:

  • You shouldn't write $(\sqrt{x} - 1)/1-1$ if you mean $(\sqrt{x} - 1)/(1-1)$; remember the conventions on order of operations.
  • If you put $1$ in place of $x$ in $(\sqrt{x} - 1)/(x-1)$, what you get is $(\sqrt{1} - 1)/(1-1)$. This is $0/0$.
  • The expression $(\sqrt{x} - 1)/(x-1)$ gives the slope of a secant line, not of a tangent line.
  • Since $0/0$ is undefined, in order to find the slope of the tangent line, you need to find $\lim\limits_{x\to1} (\sqrt{x} - 1)/(x-1)$, rather than simply plugging in $1$ in place of $x$.
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    My biggest problem is I don't know what f(x) is and what f(a) is, is is y and x or what? My book does not explain this anywhere or even show work for anything really, they just jump to the answer. Every example the book uses a = 1 but the points are always 1,1. Whoever made this book is an idiot, there is no definition of 1 in this book. Is is either x or y.2011-09-11
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y=root of x, or simply f(x)= root of x at the point (1,1) by deriving the root of x we get x^1/2 which is equal to 1/2.root of x plug in the x value from (1,1) and get 1/2 we know y,m,and x so we only need to find c in this equation, y=mx+c then we get 1=1/2*1+c c=1/2 so our equation will be y=1/2x+1/2