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)

- NathanJHW

- katieb

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- NathanJHW

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.

- NathanJHW

- jim_thompson5910

we're told that f(-1) = 1
so that means (-1,1) lies on the graph of f(x)

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## More answers

- jim_thompson5910

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

- NathanJHW

-1?

- NathanJHW

-2 sorry

- jim_thompson5910

nope on either one

- NathanJHW

How do I do it then?

- jim_thompson5910

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

- NathanJHW

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

- jim_thompson5910

oh wait, I divided on accident for some reason

- jim_thompson5910

idk why I did that

- jim_thompson5910

yeah you're right, m = -2

- jim_thompson5910

anyways, we know the following
x = -1
y = 1
m = -2
now plug those into
y = mx+b
and solve for b

- NathanJHW

b= -1

- jim_thompson5910

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

- NathanJHW

Alright, now how do we do B?

- jim_thompson5910

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

- NathanJHW

-2.8

- jim_thompson5910

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

- jim_thompson5910

-2.8? you sure?

- NathanJHW

-2(0.9)-1

- jim_thompson5910

negative 0.9

- NathanJHW

oops

- NathanJHW

0.8

- jim_thompson5910

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

- NathanJHW

Now how do we do part C?

- jim_thompson5910

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

- NathanJHW

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

- jim_thompson5910

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

- jim_thompson5910

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

- NathanJHW

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

- NathanJHW

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

- jim_thompson5910

yeah hopefully you see how it all leads up to that

- NathanJHW

I got infinity

- jim_thompson5910

me too

- NathanJHW

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

- jim_thompson5910

np

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