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Find the equation of the tangent to the curve at te given point:

B) $f(x)=\frac{1}{\sqrt(x-1)}$ At $(2,1)$

I know the procedure on how to do it but I'm getting the wrong answer. First I found the slope and had to use binomial conjugate step, etc.

The answer is $x+2y-4=0$

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    $\begingroup$ OK, so what did you get for the derivative and the line's slope? $\endgroup$ – imranfat Feb 11 '14 at 16:41
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The tangent of the curve of a function $f$ at a point $(x_0,f(x_0))$ has the equation: $$y=f'(x_0)(x-x_0)+f(x_0)$$ and in your example $f(x)=(x-1)^{-1/2}$ and $x_0=2$ hence $f'(2)=-\frac12$ so the equation is $$y=-\frac 1 2(x-2)+1\iff x+2y-4=0$$

Edited The slope of the tangent at $(2,1)$ is $$\lim_{h\to0}\frac{f(h+2)-f(2)}{h}=\lim_{h\to0}\frac{\frac{1}{\sqrt{h+1}}-1}{h}=\lim_{h\to0}\frac{(1-\sqrt{h+1})\color{red}{\times(1+\sqrt{h+1})}}{h\sqrt{h+1}\color{red}{\times(1+\sqrt{h+1})}}\\ =\lim_{h\to0}\frac{-1}{\sqrt{h+1}\color{red}{\times(1+\sqrt{h+1})}} =-\frac12$$

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  • $\begingroup$ The formula is slope of tangent= (f(a+h)-f(a))/h $\endgroup$ – user127966 Feb 11 '14 at 16:46
  • $\begingroup$ And I plugged in the x value into the equation and solved and got -h^2+4h / h. Idk what to do next $\endgroup$ – user127966 Feb 11 '14 at 16:47
  • $\begingroup$ Sorry but I don't understand you. Can you explain more? $\endgroup$ – user63181 Feb 11 '14 at 16:49
  • $\begingroup$ Sorry @Sami I do not understand what you are doing. I just learned this today as I just started grade 12 calculus. Thanks for the help can you explain in more simpler/detailed terms? $\endgroup$ – user127966 Feb 11 '14 at 16:50
  • $\begingroup$ Do you know the derivative of a function? $\endgroup$ – user63181 Feb 11 '14 at 16:51
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By differentiation we get:

$f'(x)=\frac{-(x-1)^{-3/2}}{2}$

We can then substitute $x=2$ to find the gradient of the funtion at $(2,1)$,

$f(2)=\frac{-(2-1)^{-3/2}}{2}=\frac{-1}{2}$

So the gradient is $\frac{-1}{2}$.

Now we know the slope and a point on the line, therefore we can calculate the equation of the line. The equation of a straight line is always in the form $y=mx +c$, using this fact and substituting in the gradient we just found we get:

$y=\frac{-1}{2}x+c$, luckily we also know a point on the line to substitute in to find $c$.

Using $(2,1)$:

$1=\frac{-1}{2}2+c$
rearranges to $c=2$ so, $y=\frac{-1}{2}x+2$
Multiplying everything by $2$ will give us: $2y=x+4$ or alternatively; $x+2y-4=0$

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