I am suppose to find the equation of a tangent line and I am given the following information: $ y = \sqrt[4] {x}$, $(1,1)$ I know the formula $f(a+h)-f(a)$ but it does not seem to be helping me.
Finding the equation of a tangent line
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3The equation of the tangent line to the graph of $f(x)$ at $(a,f(a))$ is $$y=f(a)+f'(a)(x-a),$$ where $f'(a)$ is the derivative of $f(x)$ at $x=a$. By definition this derivative is $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$ In your case $f(x)=\sqrt[4]{x}$ and $a=1$. – 2011-09-18
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0Shouldn't x-a be zero? – 2011-09-18
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1$x-a=h$ tends to $0$. – 2011-09-18
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0$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ – 2011-09-18
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0I am a little confused as to what I need to do because it is actually not even in this book that I can see. So I need to first find the derivative, then plug in the x and then add that into the slope for the point slope form? – 2011-09-18
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0@Jordan That's exactly what you need to do. – 2011-09-18
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0So you have already learned the rules for computing derivatives, don't you? In your question [63684](http://math.stackexchange.com/q/63684/752) you had not. As Srivatsan Narayanan commented you are right. – 2011-09-18
2 Answers
Equation of tangent line at point $(a,f(a))$ is $y = f(a) + f'(a)(x - a)$, so we have to find $f'(x)$ and than plug in value $a$ into the result.
$$ f'(x) = (\sqrt[4]{x})' = ({x^{\frac{1}{4}}})' = \frac{1}{4}{x^{\frac{{ - 3}}{4}}}$$
$$\implies f'(a) = f'(1) = \frac{1}{4}$$
Since $f(1) = 1$, we can write next equation $y = 1 + \frac{1}{4}(x - 1)$
which means that equation of tangent line is $y = \frac{1}{4}x + \frac{3}{4}$.
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0Why this impulse you have of posting the entire answer as a $\LaTeX$ equation in sundry ways? It also seems to screw up the search: I tried searching for "Equotion" (the misspelling you have twice), but search could not find it. By contrast, it was able to find the response by mixedmath when searching for "SET OF HINTS". So, could you please stop it? – 2011-09-18
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1And now that I have edited it, search finds the reply when I search for "Equotion". I think that settles it: it is activiely detrimental to the site for you to post your entire answer, text and all, as a $\LaTeX$ equation using `\text`; it makes your answers unsearchable. – 2011-09-18
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0pedja, I assumed that you somehow misunderstood that the entire equation should be posted as a $\LaTeX$ equation. So I edited it. (Later I found that @Arturo had also edited it before me.) – 2011-09-18
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0@ArturoMagidin,I am using MathType editor so that is probably reason of such LATex conversion.If you think that such formating cause problems I will stop – 2011-09-18
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0@pedja: You are posting your text *as* LaTeX equation, using `\text`. (To make it worse, you also using it haphazardly on variable names, so that sometimes you typeset $\text{m}$ and sometimes $m$, sometimes $x$ and sometimes $\text{x}$. Yes, it causes problems, and having the entire post be a bunch of $\LaTeX$ processed by MathJax makes it "invisible" to the search engine. It means nobody can find the post by searching for content. So, yes, please stop. – 2011-09-18
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0@ArturoMagidin,I will consider your suggestion :)....ok,I will stop to use such ugly formatting... – 2011-09-18
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0This isn't very clear what is happening to me are you using point slope form or what? From what I can tell the problem was in the form of y-y1 = m(x-x1) where x1 and y1 are the known values. – 2011-09-25
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0@JordanCarlyon,according to the text of the question tangent line is touching the graph at the point $(1,1)$ so that point has to satisfy tangent line equation,otherwise you can't solve the problem – 2011-09-25
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0I don't quite understand what forumla was used and why. My thought process on these problems is that I have to use the slope formula y=mx+b and then find two different point for the graph and then I could use the other formula. – 2011-09-25
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0@JordanCarlyon,try to read this carefully.. http://en.wikipedia.org/wiki/Tangent#More_rigorous_description – 2011-09-25
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0I have never seen that formula before and we were taught a different way in class. Is there not an easier way to do it than my memorizing another formula? – 2011-09-25
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0@JordanCarlyon,with this kind of given informations I don't know the other way to find a tangent line – 2011-09-26
SET OF HINTS:
For the point-slope form of a tangent lint, $y - y_0 = m (x - x_0)$, you need exactly 2 pieces of information: the slope $m$ of the line and a point $(x_0, y_0)$ that the line goes through.
You have the point, so how do you find the slope? Well, the derivative tells you something about the slope, right? If you know what that is, then you can assemble both of these pieces of information into the equation for the tangent line.
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0@Jordan: every equation for a line that I know has an x and a y in it (this is understatement- they all have variables). No? – 2011-09-18
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0y-y1=m(x-x1) I am just not sure how to use it here. – 2011-09-18
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0@Jordan: Well, then I suppose you should look into it *more*. – 2011-09-18
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0I mean I know that the equation is y-1=deriviative (x-1) but I can't get the right answer from that. – 2011-09-18
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0@Jordan: does this presuppose that you know the correct answer? – 2011-09-18
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1I have the correct answer in the back of my book, but not the ability to get to it on my own yet. – 2011-09-18