Question

1. find the equation of the tangent line to f(x)= 3x+1, x=4
which I got the first part to find P (4,13)

2. find the equation of the tangent line f(x)= sqrt (x) , x= 1/9 which I got the first part P ( 1/9, 1/3)

3. find the equation to the tangent line f(x)= 1/sqrt(x), x=1 .Again got P(1, 1)

Answer 1

can you find the 1st derivative in question 1..?

Answer 2

3?

Answer 3

and for 2. I have 1 / 2 sqrt(x)
and for 3 I have 1/(2*x^(3/2))

Answer 4

ok... so for 2 substitute x = 1/9 will give the slope at x = 1/9

then find the equation of the straight line
..

for 3 substitute x = 1 and this will be the slope of the tangent... then you can use the point to find the equation of the line...

Answer 5

for the 1st question the slope is 3 and the point is (4, 13)

you'll see something interesting when you find the equation of the tangent.

Answer 6

for question 2 I got y= 3/2x + 3/18

Answer 7

using y = 3/2x + b

1/3 = 3/2(1/9) + b

so

1/3 = 1/6 + b

so b = 1/6

...

Answer 8

question 1 is 3x+1
which is the same from the original problem

Answer 9

and actually there is another tangent in question 2, where m = -3/2

so y = -3/2 x + b

1/3 = -3/2(1/9) + b

1/2 = b

Answer 10

thats correct...

Answer 11

but thats interesting since a tangent only touches at 1 point...

the equation you have touches at all points..

Answer 12

in question 2

\[\sqrt{\frac{1}{9}} = \pm \frac{1}{3} \]

Answer 13

I think in question 3 your 1st derivative is incorrect

\[y = \frac{1}{\sqrt{x} }.... or......y = x^{-\frac{1}{2}}\]

Answer 14

so

\[\frac{dy}{dx} = - \frac{1}{2} x^{-\frac{3}{2}} ... or ... \frac{dy}{dx} = -\frac{ 1}{2\sqrt{x^3}}\]

Answer 15

so again 2 tangents... as you have 2 slopes...

Answer 16

I get y= -1/2x + 3/2 for the equation

Answer 17

that looks good...

I have to go... hope I helped...

Answer 18

thank you