What is the tangent line to $y=e^{^{\frac{x}{2}}}$ through (0,0)?

Question

I'm trying to solve

$y=e^{^{\frac{x}{2}}}$

The derivative:

$\frac{\sqrt{e^{^x}}}{2}$

So, I need to find the "slope" to the linear function $y\:-\:y_1\:=m\left(x-x_1\right)$, solving the derivative by replacing $x$ by $0$ is $m=\frac{1}{2}$, so the answer is:

$y-y_1=m\left(x-x_1\right),\:y-0=\frac{1}{2}\left(x-0\right),\:y=\frac{1}{2}x$

But the answer that Wolfram Alpha gives me is:

$\frac{ex}{2}$

So, does this problem requires another formula or process to be solved? Or did I just fail in the process?

Greetings!

Feel free to edit the post if there are any English issues in the post, I appreciate it so much!

If you put $x=0 $ in $f'=\frac12 e^{\frac{x}{2}} $ slope is $\frac{1}{2}e^0=\frac12 $ — Khosrotash, Aug 02 '17 at 16:55
Anyone else get the strong feeling answerers will probably misread the entire question? Perhaps the WA tag might ring some bells as to what seems to be the real question. — Simply Beautiful Art, Aug 02 '17 at 17:00
@SimplyBeautifulArt I don't understand your comment. My answer agrees with WA. And it is the correct way to solve the question as it is formulated. — mfl, Aug 02 '17 at 17:05
More so I meant to point out that the problem of WA's result isn't really emphasized, more so glazed over. It'd probably be helpful if the OP more specifically asked why WA gave that result, unless I am misinterpreting the question. — Simply Beautiful Art, Aug 02 '17 at 17:06
@SimplyBeautifulArt https://www.wolframalpha.com/input/?i=tangent+line+to+y%3De%5E(x%2F2)+through+(0,0) gives the solution. You don't need to say sorry. I only asked for clarification. — mfl, Aug 02 '17 at 17:08
Huh. I think I need new glasses. Somehow I read the WA output as $\frac{cx}2\ne\frac{ex}2$... — Simply Beautiful Art, Aug 02 '17 at 17:14
@SimplyBeautifulArt I don't think so. I have read the same. Unless I need new glasses too. — mfl, Aug 02 '17 at 17:18
@SimplyBeautifulArt I'm curious about what do you mean with "WA" ^^ — Andrés David Hernández Sánchez, Aug 02 '17 at 17:19
xD Us blind people staring at our computer screens all day >.> — Simply Beautiful Art, Aug 02 '17 at 17:19

score 3 · Accepted Answer · answered Aug 02 '17 at 16:57

3

Firs of all note that $(0,0)$ is not a point of the graph of the function. So fix $x_0\in\mathbb{R}.$ The tangent line at $(x_0,y_0)$ is given by $$y-e^{x_0/2}=\frac 12 e^{x_0/2}(x-x_0).$$ (Note that $f'(x)=\frac12 e^{x/2}$.) If this line contains the point $(0,0)$ then we have $$e^{x_0/2}=\frac 12 e^{x_0/2}x_0.$$ Since $e^{x_0/2}\ne 0$ we get that $x_0=2.$ Thus the tangent line is

$$y-e=\frac{e}{2}(x-2)$$ That is $$y=\frac{e}{2}x.$$

answered Aug 02 '17 at 16:57

mfl

29,399

So many thanks @mfl! I have a question..
Why this $e^{^{\frac{x_0}{2}}}\ne 0$? And why $x_0=2$?
– Andrés David Hernández Sánchez Aug 02 '17 at 17:08
@AndrésDavidHernándezSánchez $x_0=2$ was just an example. You didn't specify which point you wanted to take the tangent line at in WA probably. – Simply Beautiful Art Aug 02 '17 at 17:09
We have that $e^x>0$ for all real $x.$ So it is $\ne 0.$ Thus we can divide both terms of the equality $e^{x_0/2}=\frac 12 e^{x_0/2}x_0$ by the nonzero number $e^{x_0/2}.$ Thus we get $1=\frac12 x_0.$ That is, $x_0=2.$ – mfl Aug 02 '17 at 17:09
@SimplyBeautifulArt You don't need to say the point where you take the tangent. See https://www.wolframalpha.com/input/?i=tangent+line+to+y%3De%5E(x%2F2)+through+(0,0) – mfl Aug 02 '17 at 17:12
No everything have sence! So many thanks @mfl, I never expect to do this analisys for $e^{^x}>0$... – Andrés David Hernández Sánchez Aug 02 '17 at 17:16
You're welcome. Remember that if $f(x)=e^{x/2}$ then it is $f'(x)=\frac12 e^{x/2}\ne \frac{\sqrt{e^x}}{2}.$ – mfl Aug 02 '17 at 17:21

score 1 · Answer 2 · answered Aug 02 '17 at 16:53

1

How did you take the derivative here? Remember that $\frac{d}{dx}(e^x) = e^x$ and try using the chain rule.

answered Aug 02 '17 at 16:53

platty

3,555

What is the tangent line to $y=e^{^{\frac{x}{2}}}$ through (0,0)?

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