0
$\begingroup$

This question already has an answer here:

Use the definition of the derivative to find $f'(x)$ for $f(x)=\sqrt{x-2}$. The answer that I got was $$\frac{1}{2(x-2)^.5}$$. Is this correct? The second part asks use your answer from part 1 to find the equation of the line tangent to $f(x)=\sqrt{x-2}$ at the point $(6,2)$. How do I do this?

$\endgroup$

marked as duplicate by user147263, daw, Adam Hughes, Tomás, egreg Sep 22 '14 at 21:25

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

1
$\begingroup$

Your result for $f'(x)$ is correct.

For the second part you need to find the line defined by

$$y = mx + b$$

$m$ is the slope of the line and is simply $f'(6)$ since the line must be tangent to $f(x)$ at $x = 6$ (the tangent has the same slope). Now you need to solve for $b$. Plug in $x = 6$ and $y = 2$ into $y = f'(6)x + b$ and solve for $b$.

$\endgroup$
2
$\begingroup$

The definition of the derivative is:

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

Using $f(x)=\sqrt{x-2}$, $$\begin{array}{rcl} f'(x)&=&\lim_{h\to0}\frac{\sqrt{x+h-2}-\sqrt{x-2}}{h}\\ &=&\lim_{h\to0}\frac{\sqrt{x+h-2}-\sqrt{x-2}}{h}\frac{\sqrt{x+h-2}+\sqrt{x-2}}{\sqrt{x+h-2}+\sqrt{x-2}}\\ &=&\lim_{h\to0}\frac{(x+h-2)-(x-2)}{h}\frac{1}{\sqrt{x+h-2}+\sqrt{x-2}}\\ &=&\lim_{h\to0}\frac{h}{h}\frac{1}{\sqrt{x+h-2}+\sqrt{x-2}}\\ &=&\lim_{h\to0}\frac{1}{\sqrt{x+h-2}+\sqrt{x-2}}\\ &=&\frac{1}{\sqrt{x-2}+\sqrt{x-2}}\\ \Rightarrow f'(x)&=&\frac{1}{2\sqrt{x-2}}\\ \end{array}$$

Then,note that the derivative evaluated at $x_0$ is the slope of the tangent line of the function $f$ at $x_0$, that is, $$\text{tangent line at }x_0:\,y-y_0=f'(x_0)(x-x_0)$$

So $$\begin{array}{rcl} y-2&=&f'(6)\big(x-6\big)\\ \Rightarrow y&=&\frac{1}{2\cdot\sqrt{6-2}}(x-6)+2\\ \Rightarrow y&=&\frac{1}{4}(x-6)+2\\ \Rightarrow y&=&\frac{1}{4}x +\frac{1}{2}\\ \end{array}$$

$\endgroup$
0
$\begingroup$

I'm not entirely sure of the first answer. It would help to use TeX to display it. Do you mean:

$\frac{1}{2(x-2)^{.5}}$ or $\frac{1}{2}(x-2)^{.5}$ ?

(The first one is correct). As for the second part, the line tangent to $f(x)$ at (6,2), is going to be a straight line with a slope equal to the derivative of $f(x)$ at that point (which you found before), that passes through the point (6,2). So find the value of the derivative at $x=6$, and that value is the slope. Then use an equation for a line (such as $y-y_{1}=m(x-x_{1})$ where $m$ is the slope, and $y_{1}$ and $x_{1}$ correspond to the point values.)

$\endgroup$
0
$\begingroup$

$f^{`}(x)=\frac {1}{2 \sqrt{x-2}}$ then $$m_{t} = \frac {1}{2\sqrt{6-2}} = \frac{1}{4}$$

you can find the equation of any line with $y-y_{0} = m (x-x_{0})$

so the equation of the tangent line is $y-2 = \frac {1}{4} (x-6)$ and its done.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.