find the equation of the tangent line to the curve $y = {\sqrt{x+2}}$ at the point where $ x=2$

Question

The question asks for the equation of the tangent line.

So I need this formula $$y-y_1 =m(x-x_1)$$

and the concept of derivatives

$$\lim_{Δ x\to 0} \frac{f(x+Δx) - f(x)}{Δx}$$

Ok, now I have that, I need to find $f(2)$. Let's proceed.

$$f(2) = {\sqrt{2+2}} = 2$$

Wait, But $\sqrt{4}$ also has $-2$ as a root, what should I do? go on with $2$ or with $-2$?

I intended to find the slope and then plug it in the formula with the point $(2,2)$ to find the equation. If someone can solve this question I'll be grateful.

$\sqrt{x}$ is only the positive root. It is also called the principal root. So it would be 2. — user388557, Jul 12 '17 at 21:36
Not necessarily. The best thing to do is ask the person who posed the question what exactly they meant. — NickD, Jul 12 '17 at 21:40
@Nick No, not really. $\sqrt x$ does imply the positive root only. — Sawarnik, Jul 12 '17 at 21:42
For the reason why we choose the positive root, see Why is $\sqrt{x}$ a function? — N. F. Taussig, Jul 12 '17 at 21:51

Arnaldo · Accepted Answer · 2017-07-12T21:46:47.350

2

Hint

$$y'(x)=\frac{1}{2\sqrt{x+2}}\to m=y'(2)=\frac{1}{4}$$

and $(x_1,y_1)=(2,f(2))=(2,2)$.

Ps.: By definition, $\sqrt{4}=2$, and more general, if we have $a>0$ then $\sqrt{a}>0.$

Can you finish?

edited Jul 12 '17 at 21:46

answered Jul 12 '17 at 21:38

Arnaldo

21,342

Yes I had found this result, But I didn't understand why You choose 2 instead of -2. – Goun2 Jul 12 '17 at 21:41
@hjx: because, by definition, $\sqrt{4}=2$ – Arnaldo Jul 12 '17 at 21:44
1

Ok then, Valeu Arnaldo. – Goun2 Jul 12 '17 at 21:47

find the equation of the tangent line to the curve $y = {\sqrt{x+2}}$ at the point where $ x=2$

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