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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.

hmakholm left over Monica
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    The 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 $a=1$. – Américo Tavares Sep 18 '11 at 00:56
  • Shouldn't x-a be zero? –  Sep 18 '11 at 01:16
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    $x-a=h$ tends to $0$. – Américo Tavares Sep 18 '11 at 01:36
  • $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$$ – Américo Tavares Sep 18 '11 at 01:40
  • I 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? –  Sep 18 '11 at 01:45
  • @Jordan That's exactly what you need to do. – Srivatsan Sep 18 '11 at 04:00
  • So you have already learned the rules for computing derivatives, don't you? In your question 63684 you had not. As Srivatsan Narayanan commented you are right. – Américo Tavares Sep 18 '11 at 10:47

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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 $y = \frac{1}{4}x + \frac{3}{4}$.

Srivatsan
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Pedja
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  • Why 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? – Arturo Magidin Sep 18 '11 at 04:01
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    And 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. – Arturo Magidin Sep 18 '11 at 04:05
  • pedja, 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.) – Srivatsan Sep 18 '11 at 04:07
  • @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 – Pedja Sep 18 '11 at 04:09
  • @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. – Arturo Magidin Sep 18 '11 at 04:16
  • @ArturoMagidin,I will consider your suggestion :)....ok,I will stop to use such ugly formatting... – Pedja Sep 18 '11 at 04:22
  • This 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. –  Sep 25 '11 at 18:01
  • @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 – Pedja Sep 25 '11 at 18:41
  • I 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. –  Sep 25 '11 at 19:54
  • @JordanCarlyon,try to read this carefully.. http://en.wikipedia.org/wiki/Tangent#More_rigorous_description – Pedja Sep 25 '11 at 20:14
  • I 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? –  Sep 25 '11 at 23:00
  • @JordanCarlyon,with this kind of given informations I don't know the other way to find a tangent line – Pedja Sep 26 '11 at 05:26
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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.

davidlowryduda
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