## NathanJHW one year ago 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)

1. 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.

2. NathanJHW

@jim_thompson5910

3. jim_thompson5910

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

4. jim_thompson5910

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

5. NathanJHW

-1?

6. NathanJHW

-2 sorry

7. jim_thompson5910

nope on either one

8. NathanJHW

How do I do it then?

9. jim_thompson5910

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

10. NathanJHW

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

11. jim_thompson5910

oh wait, I divided on accident for some reason

12. jim_thompson5910

idk why I did that

13. jim_thompson5910

yeah you're right, m = -2

14. jim_thompson5910

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

15. NathanJHW

b= -1

16. jim_thompson5910

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

17. NathanJHW

Alright, now how do we do B?

18. jim_thompson5910

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

19. NathanJHW

-2.8

20. 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

21. jim_thompson5910

-2.8? you sure?

22. NathanJHW

-2(0.9)-1

23. jim_thompson5910

negative 0.9

24. NathanJHW

oops

25. NathanJHW

0.8

26. jim_thompson5910

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

27. NathanJHW

Now how do we do part C?

28. 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))

29. NathanJHW

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

30. 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 $\Large |y+1| = 2*e^{x^2+x}$

31. 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 $\Large |y+1| = 2*e^{x^2+x}$ $\Large y+1 = 2*e^{x^2+x}$ $\Large y = 2*e^{x^2+x}-1$ $\Large f(x) = 2*e^{x^2+x}-1$

32. NathanJHW

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

33. NathanJHW

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

34. jim_thompson5910

yeah hopefully you see how it all leads up to that

35. NathanJHW

I got infinity

36. jim_thompson5910

me too

37. NathanJHW

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

38. jim_thompson5910

np