Answer: (a) \(x=-1\), \(y=x\)

(b) \((0,1)\) and \((-2,-3)\)

(c) Sketch with vertical asymptote \(x=-1\), oblique asymptote \(y=x\), stationary points \((0,1)\) and \((-2,-3)\), and correct branch behaviour.

(d) Take the right branch of part (c) and reflect it in the \(y\)-axis; the graph is symmetric about \(x=0\) and smooth at \((0,1)\).

(e) \(-1-\sqrt3

(a)

The denominator is zero when \(x=-1\), so \(x=-1\) is the vertical asymptote.

Divide:

\(\frac{x^2+x+1}{x+1}=x+\frac1{x+1}.\)

As \(|x|\to\infty\), \(\frac1{x+1}\to0\), so the oblique asymptote is

\(y=x.\)

(b)

Write

\(y=x+\frac1{x+1}.\)

Then

\(\frac{dy}{dx}=1-\frac1{(x+1)^2}=\frac{x^2+2x}{(x+1)^2}.\)

Set \(\frac{dy}{dx}=0\):

\(x^2+2x=x(x+2)=0.\)

So \(x=0\) or \(x=-2\). The corresponding values are

\(y(0)=1,\qquad y(-2)=-3.\)

Therefore the stationary points are \((0,1)\) and \((-2,-3)\).

(c)

The curve can be written as

\(y=x+\frac1{x+1}.\)

So the asymptotes are \(x=-1\) and \(y=x\).

The right branch, for \(x>-1\), approaches \(+\infty\) as \(x\to-1^+\), has a smooth minimum at \((0,1)\), then approaches \(y=x\) from above.

The left branch, for \(x<-1\), approaches \(y=x\) from below as \(x\to-\infty\), has a smooth maximum at \((-2,-3)\), then tends to \(-\infty\) as \(x\to-1^-\).

(d)

Let \(t=|x|\). Then

\(y=\frac{t^2+t+1}{t+1},\qquad t\ge0.\)

This is the same as the right-hand branch of \(C\) from part (c), reflected in the \(y\)-axis. The graph is symmetric about \(x=0\), has a smooth minimum at \((0,1)\), and approaches \(y=|x|\) for large \(|x|\).

(e)

Let \(t=|x|\), where \(t\ge0\). The inequality becomes

\(\frac{t^2+t+1}{t+1}<3.\)

Since \(t+1>0\), multiply through:

\(t^2+t+1<3t+3.\)

So

\(t^2-2t-2<0.\)

The roots are

\(t=1\pm\sqrt3.\)

Since \(t\ge0\), the relevant critical value is \(t=1+\sqrt3\). Therefore

\(|x|<1+\sqrt3,\)

so

\(-1-\sqrt3

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