1. ## derivitives

For 1) and 2), Find the equation for the tangent line to the graph of the function . I got another one right in point-slope but these, i am pretty lost on. 1) f(x)= s^3 at x=1/2 ans.3x-4y+1=0. 2)f(x)= 1/(x+3) at x=2. ans. x+25y-7=0 This one is for the normal line. 3) f(x)=1/(x+3) at x=3. ans 216x-6y-647=0 On a side note, I do know the equation for derivatives (f(x+change of x)-f(x))/change of x and the point-slope form is applied somewhere. I just don't know how to get these to that form.

2. Originally Posted by driver327
For 1) and 2), Find the equation for the tangent line to the graph of the function . I got another one right in point-slope but these, i am pretty lost on. 1) f(x)= s^3 at x=1/2 ans.3x-4y+1=0. 2)f(x)= 1/(x+3) at x=2. ans. x+25y-7=0 This one is for the normal line. 3) f(x)=1/(x+3) at x=3. ans 216x-6y-647=0 On a side note, I do know the equation for derivatives (f(x+change of x)-f(x))/change of x and the point-slope form is applied somewhere. I just don't know how to get these to that form.
(It still doesn't look right)

Anyway,

Find the equation for the tangent line to $f(x)= x^3$ at $x= \frac {1}{2}$
The derivative gives the slope of the tangent line, so let's find it first.

$f(x) = x^3$

$\Rightarrow f'(x) = 3x^2$

At $x = \frac {1}{2}$, we get:

$f' \left( \frac {1}{2} \right) = 3 \left( \frac {1}{2} \right)^2 = \frac {3}{4}$

Now let's find the point we want to work on.

when $x = \frac {1}{2}$, $y = \left( \frac{1}{2} \right)^3 = \frac {1}{8}$

So using the $m = \frac {3}{4}$ and $(x_1, y_1) = \left( \frac {1}{2}, \frac {1}{8} \right)$

We have by the point-slope form:

$y - y_1 = m(x - x_1)$

$\Rightarrow y - \frac {1}{8} = \frac {3}{4} \left( x - \frac {1}{2} \right)$

$\Rightarrow y = \frac {3}{4}x - \frac {1}{4}$

$\Rightarrow - \frac {3}{4} x + y + \frac {1}{4} = 0$

$\Rightarrow 3x - 4y - 1 = 0$

The answer you gave is incorrect. That would be the answer if we wanted it at $x = - \frac {1}{2}$

(2) Find the equation for the tangent line to f(x)= 1/(x+3) at x=2. ans. x+25y-7=0

3) Find the normal line to f(x)=1/(x+3) at x=3. ans 216x-6y-647=0
I leave these to you, they are very similar. The difference with the normal line is that it is perpendicular to the tangent line. So when you find the slope of the tangent line, take it's negative inverse and use that as the slope for the normal line

3. (2) Find the equation for the tangent line to f(x)= 1/(x+3) at x=2. ans. x+25y-7=0
$f(x) = \frac{1}{(x+3)}$

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

$Set \ x = 2 \ in \ \ \frac{-1}{(x+3)^2}$

$Thus \ m = \frac{-1}{25}$

$f(x) = \frac{1}{(x+3)}$
$Set \ x = 2 \ to \ get \ y's \ value \ \ f(x) = \frac{1}{(x+3)}$

$Thus \ y = \frac{1}{5} \ where \ x = 2$

$y = mx + c$

$\frac{1}{5} = ( \frac{-1}{25} ) ( 2 ) + c$

$c = \frac{7}{25}$

$y = \frac{-1}{25} x + \frac{7}{25}$

$0 = 25y + x - 7$

4. Find the normal line to f(x)=1/(x+3) at x=3. ans 216x-6y-647=0
$f(x) = \frac{1}{(x+3)}$

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

$Set \ x = 3 \ in \ \ \frac{-1}{(3+3)^2}$

$Thus \ m = \frac{-1}{36}$

BUT: They ask us for the NORMAL. So we have to multiply the gradient with $-1$

$Thus \ m = 36$

$f(x) = \frac{1}{(x+3)}$
$Set \ x = 3 \ to \ get \ y's \ value \ \ f(x) = \frac{1}{(3+3)}$

$Thus \ y = \frac{1}{6} \ where \ x = 3$

$y = mx + c$

$\frac{1}{6} = ( 36 ) ( 3 ) + c$

$c = -108 + \frac{1}{6}$

$y = 36 x + (-108 + \frac{1}{6})$

5. Originally Posted by janvdl

$y = -36 x + (-108 + \frac{1}{6})$

Something doesnt seem right.
Can someone check this one for me please?
One mistake Jan. The 36 in front of x is positive.

$y = 36x + \left( -108 + \frac {1}{6} \right)$

$\Rightarrow y = 36x - \frac {647}{6}$

$\Rightarrow 36x - y - \frac {647}{6} = 0$

$\Rightarrow 216x - 6y - 647 = 0$

as desired

6. Thanks Jhevon, i fixed it