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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)
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)

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NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0A. 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.

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2we're told that f(1) = 1 so that means (1,1) lies on the graph of f(x)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2plug in x = 1 and y = 1 into the dy/dx formula and that will determine the slope m

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2nope on either one

NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0How do I do it then?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2you plugged it into dy/dx = (2x + 1)( y + 1) right?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2oh wait, I divided on accident for some reason

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2idk why I did that

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yeah you're right, m = 2

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2anyways, we know the following x = 1 y = 1 m = 2 now plug those into y = mx+b and solve for b

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2so the tangent line is y = 2x  1

NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0Alright, now how do we do B?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2plug in x = 0.9 into the tangent line equation to get the approximate value for f(0.9)

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2the 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
 one year ago
Best ResponseYou've already chosen the best response.22.8? you sure?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2that's the approx value of f(0.9)

NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0Now how do we do part C?

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \frac{dy}{dx} = (2x+1)(y+1)\] \[\Large dy = (2x+1)(y+1)dx\] \[\Large \frac{dy}{y+1} = (2x+1)dx\] \[\Large \int\frac{dy}{y+1} = \int(2x+1)dx\] \[\Large \ln(y+1) = x^2+x+C\] I'll let you finish up. The goal is to isolate y (which is the same as finding f(x))

NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0Is it y = e^(C+x^2+x)1

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2\[\Large \ln(y+1) = x^2+x+C\] \[\Large y+1 = e^{x^2+x+C}\] \[\Large y+1 = e^C*e^{x^2+x}\] 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
 one year ago
Best ResponseYou've already chosen the best response.2The 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
 one year ago
Best ResponseYou've already chosen the best response.0Alright, sorry I had to go somewhere @jim_thompson5910 back now though.

NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0For the last one, f(x) is the answer to c right? @jim_thompson5910

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.2yeah hopefully you see how it all leads up to that

NathanJHW
 one year ago
Best ResponseYou've already chosen the best response.0alright thank you so much for all your help @jim_thompson5910 really appreciate it!
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