NathanJHW
  • NathanJHW
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)
Mathematics
katieb
  • katieb
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NathanJHW
  • 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
  • NathanJHW
jim_thompson5910
  • 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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jim_thompson5910
  • jim_thompson5910
plug in x = -1 and y = 1 into the dy/dx formula and that will determine the slope m
NathanJHW
  • NathanJHW
-1?
NathanJHW
  • NathanJHW
-2 sorry
jim_thompson5910
  • jim_thompson5910
nope on either one
NathanJHW
  • NathanJHW
How do I do it then?
jim_thompson5910
  • jim_thompson5910
you plugged it into dy/dx = (2x + 1)( y + 1) right?
NathanJHW
  • NathanJHW
(2(-1)+1)(1+1)
jim_thompson5910
  • jim_thompson5910
oh wait, I divided on accident for some reason
jim_thompson5910
  • jim_thompson5910
idk why I did that
jim_thompson5910
  • jim_thompson5910
yeah you're right, m = -2
jim_thompson5910
  • 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
  • NathanJHW
b= -1
jim_thompson5910
  • jim_thompson5910
so the tangent line is y = -2x - 1
NathanJHW
  • NathanJHW
Alright, now how do we do B?
jim_thompson5910
  • jim_thompson5910
plug in x = -0.9 into the tangent line equation to get the approximate value for f(-0.9)
NathanJHW
  • NathanJHW
-2.8
jim_thompson5910
  • 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
  • jim_thompson5910
-2.8? you sure?
NathanJHW
  • NathanJHW
-2(0.9)-1
jim_thompson5910
  • jim_thompson5910
negative 0.9
NathanJHW
  • NathanJHW
oops
NathanJHW
  • NathanJHW
0.8
jim_thompson5910
  • jim_thompson5910
that's the approx value of f(-0.9)
NathanJHW
  • NathanJHW
Now how do we do part C?
jim_thompson5910
  • 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
  • NathanJHW
Is it y = e^(C+x^2+x)-1
jim_thompson5910
  • 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
  • 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
  • NathanJHW
Alright, sorry I had to go somewhere @jim_thompson5910 back now though.
NathanJHW
  • NathanJHW
For the last one, f(x) is the answer to c right? @jim_thompson5910
jim_thompson5910
  • jim_thompson5910
yeah hopefully you see how it all leads up to that
NathanJHW
  • NathanJHW
I got infinity
jim_thompson5910
  • jim_thompson5910
me too
NathanJHW
  • NathanJHW
alright thank you so much for all your help @jim_thompson5910 really appreciate it!
jim_thompson5910
  • jim_thompson5910
np

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