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)

1. Write an equation for the line tangent to the curve y = f (x) at x = ? 1.

2. Use the tangent line from part A to estimate f (?0.9)

3. Use separation of variables to find an explicit or implicit formula for y = f (x), with no integrals remaining.

4. What is the limit of f(x) as x approaches infinity?

Answer 1

1. When $x=-1$, you know that $y=f(-1)=1$. The tangent line at $x=-1$ has slope

$\dfrac{\mathrm dy}{\mathrm dx}(-1,1)=(2(-1)+1)(1+1)=-2$

Then the tangent line has equation

$y-1=-2(x+1)\implies\boxed{y=-2x-1}$

2. Plug $x=-0.9$ into the equation for the tangent line to get

$f(-0.9)\approx-2(-0.9)-1\implies\boxed{f(-0.9)\approx0.8}$

3. Separating the variables in the ODE gives

$\dfrac{\mathrm dy}{y+1}=(2x+1)\,\mathrm dx$

Integrating both sides yields

$\ln|y+1|=x^2+x+C$

Given that $f(-1)=1$, we get

$\ln|1+1|=(-1)^2+(-1)+C\implies C=\ln2$

so that the particular solution to the ODE is

$\ln|y+1|=x^2+x+\ln2\implies y+1=e^{x^2+x+\ln 2}\implies\boxed{y=2e^{x^2+x}-1}$

4. As $x\to\infty$, the exponential terms grows without bound, so that $\boxed{f(x)\to\infty}$ as well.