Question

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)

Answer 1

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) as x approaches infinity.

Answer 2

@jim_thompson5910

Answer 3

we're told that f(-1) = 1

so that means (-1,1) lies on the graph of f(x)

Answer 4

plug in x = -1 and y = 1 into the dy/dx formula and that will determine the slope m

Answer 5

-1?

Answer 6

-2 sorry

Answer 7

nope on either one

Answer 8

How do I do it then?

Answer 9

you plugged it into dy/dx = (2x + 1)( y + 1) right?

Answer 10

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

Answer 11

oh wait, I divided on accident for some reason

Answer 12

idk why I did that

Answer 13

yeah you're right, m = -2

Answer 14

anyways, we know the following

x = -1
y = 1
m = -2

now plug those into

y = mx+b

and solve for b

Answer 15

b= -1

Answer 16

so the tangent line is y = -2x - 1

Answer 17

Alright, now how do we do B?

Answer 18

plug in x = -0.9 into the tangent line equation to get the approximate value for f(-0.9)

Answer 19

-2.8

Answer 20

the idea is that if we plug in values close to x = -1, then we'll get rough approximations to f(x)

be warned: the further you stray from x = -1, the larger the error will be until it will be so large that it's not worth it

Answer 21

-2.8? you sure?

Answer 22

-2(0.9)-1

Answer 23

negative 0.9

Answer 24

oops

Answer 25

0.8

Answer 26

that's the approx value of f(-0.9)

Answer 27

Now how do we do part C?

Answer 28

\[\Large \frac{dy}{dx} = (2x+1)(y+1)\]
\[\Large dy = (2x+1)(y+1)dx\]
\[\Large \frac{dy}{y+1} = (2x+1)dx\]
\[\Large \int\frac{dy}{y+1} = \int(2x+1)dx\]
\[\Large \ln(|y+1|) = x^2+x+C\]
I'll let you finish up. The goal is to isolate y (which is the same as finding f(x))

Answer 29

Is it y = e^(C+x^2+x)-1

Answer 30

\[\Large \ln(|y+1|) = x^2+x+C\]
\[\Large |y+1| = e^{x^2+x+C}\]
\[\Large |y+1| = e^C*e^{x^2+x}\]
Let's find the value of e^C using x = -1 and y = 1
\[\Large |y+1| = e^C*e^{x^2+x}\]
\[\Large |1+1| = e^C*e^{(-1)^2+(-1)}\]
\[\Large |2| = e^C*e^{0}\]
\[\Large 2 = e^C*1\]
\[\Large 2 = e^C\]
\[\Large e^C = 2\]
-------------------------------------------------------
So
\[\Large |y+1| = e^C*e^{x^2+x}\]
turns into
\[\Large |y+1| = 2*e^{x^2+x}\]

Answer 31

The absolute value is often a pain, but the good news is that because the right side is never negative, we can drop the absolute values.

So solving for y gives

\[\Large |y+1| = 2*e^{x^2+x}\]
\[\Large y+1 = 2*e^{x^2+x}\]
\[\Large y = 2*e^{x^2+x}-1\]
\[\Large f(x) = 2*e^{x^2+x}-1\]

Answer 32

Alright, sorry I had to go somewhere @jim_thompson5910 back now though.

Answer 33

For the last one, f(x) is the answer to c right? @jim_thompson5910

Answer 34

yeah hopefully you see how it all leads up to that

Answer 35

I got infinity

Answer 36

me too

Answer 37

alright thank you so much for all your help @jim_thompson5910 really appreciate it!

Answer 38

np