Question

Urgent Calculus help please

Answer 1

The function f (x) has the value f (−1) = 1. The slope of the curve y = f (x) at any point is given by the expression dy/dx = (2x + 1)( y + 1)

A. A. Write an equation for the line tangent to the curve y = f (x) at x = − 1.
B. Use the tangent line from part A to estimate f (−0.9)
C. Use separation of variables to find an explicit or implicit formula for y = f (x), with no integrals remaining.
D.Find lim f (x)
x-> ∞

Answer 2

@zepdrix

Answer 3

Do you have any working? Or you want to understand the question from scratch?

Answer 4

Scratch

Answer 5

Use the equation y =mx+c where m is dy/dx and y is 1 and x is -1

Answer 6

Can you do it?

Answer 7

Ok so just plug those in? so 1=dy/dx(-1)+c?

Answer 8

And you also have the expression for derivative use that too

Answer 9

1=-1(2x + 1)( y + 1)+c ok so this?

Answer 10

Plug in x and y values in the derivative too

Answer 11

Gosh dangit LOL ok I got that so solve?

Answer 12

Yeah figure out c

Answer 13

c=1??? Or am I confusing myself

Answer 14

I am getting -1

Answer 15

hmmm maybe i messed up

Answer 16

I'm still getting 1 hm

Answer 17

Post your working :-)

Answer 18

1=-1(2(-1) + 1)( 1 + 1)+c
1= 2+c
-1=c
I see my mistake now!!! oops!

Answer 19

Now your equation will be
Y=dy/dx x -1
Substitute the numerical value of dy/dx by substituting x and y value in it

Answer 20

ok

Answer 21

What do i do now

Answer 22

Still need help?

Answer 23

What's the final equation you got?

Answer 24

Not sure if i;m right

Answer 25

@baru Help?

Answer 26

y=-2 I don't think that's right @faiqRaees

Answer 27

@agent0smith

Answer 28

First use the given expression for dy/dx to find m - just plug in x=-1 and y=1 into the dy/dx

Answer 29

I think I did that already no?

Answer 30

I don't know, I'm not gonna try to read through all this to figure out where you're up to, it's too hard to follow in progress questions.

Once you have dy/dx, that's m, for y=mx+b

Answer 31

dy/dx = (2x + 1)( y + 1)

Answer 32

Yes but do as I said, plug in x and y

Answer 33

dy/dx= -2

Answer 34

No... show work.

Answer 35

(-2+1)(1+1)
(-1)(2)
-2

Answer 36

Oh sorry yes I thought it was division.

Now you have m. Now you can make the equation of a line, since you have the slope and the point the line passes through (-1, 1)

Answer 37

y=-2x+1? I don;t recall the point

Answer 38

I just gave you the point.

Answer 39

you gave me -1 and 1

Answer 40

Which is the point.

Answer 41

Ok... so I'm supposed to use them both in the equation?

Answer 42

You have the slope and a point. Use them to make the equation of a line.

Answer 43

Like y=mx+b or y-y1 =m (x-x1)

Answer 44

Oh ok so point slope y+1=-2(x-1)?

Answer 45

You made mistakes

Answer 46

y-1=-2(x+1) my bad

Answer 47

Yep

Answer 48

That's it for A?

Answer 49

Yep and part b is easy

Answer 50

Just plug in the given x value into your equation

Make a new post for part c since this is too long already.

Answer 51

So B would be y=1.2

Answer 52

and Ok

Answer 53

Looks right

Answer 54

I dont think part B is correct
Part B has to be done by the tangent line approximation formula
f(-0.9) = f(-1)+f'(-1)(-0.9+1)

Answer 55

^that's the same thing.

Answer 56

That's so not the same thing

Answer 57

What you asked her to do was find the y coordinate of the point on the tangent line when x coordinate is -0.9

Answer 58

Firstly, the question specifically asked her to use her tangent line from part a.

Secondly
f(-0.9) = f(-1)+f'(-1)(-0.9+1)
= 1 + (-2)(0.1)
= 0.8

y-1=-2(x+1)
y-1=-2(-0.9+1)
y -1 = -2(0.1)
y = 0.8

Same thing like i said. That is what the tangent line is from part a... the same thing as what you gave.

Answer 59

The tangent line approximation formula is the same thing as finding the tangent line, and using it to approximate.

Answer 60

\[\large f(x)=f(x_0)+f'(x_0)(x−x_0)\] this is the same thing as \[\large f(x)- f(x_0)=f'(x_0)(x−x_0)\] note the similarity to \[\large y- y_1 = m(x-x_1)\]

They're the exact same thing, the formula for tangent line approximation, and actually finding the tangent line at a point, and using it to approximate the value of the function

Answer 61

Thanks for clarifying that