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

`(\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

`(\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.