Find the angle at which the curve y = arctan (1+ 1/x)

Find the angle at which the curve $y = \arctan\!\left(1+\dfrac{1}{x}\right)$ crosses the $x$-axis.

Step 1: Find the crossing point

The arctangent vanishes only when its argument does: \[\arctan\!\left(1+\frac1x\right) = 0 \;\Longleftrightarrow\; 1+\frac1x = 0 \;\Longleftrightarrow\; x = -1.\]

So the curve meets the $x$-axis at the single point $(-1,\,0)$.

Step 2: Compute the slope

With $u = 1+\dfrac1x$, $u' = -\dfrac{1}{x^{2}}$: $y' = \frac{u'}{1+u^{2}} = \frac{-\dfrac{1}{x^{2}}}{1+\left(1+\dfrac1x\right)^{2}}$.

Multiplying numerator and denominator by $x^{2}$ clears the compound fraction: $y' = \frac{-1}{x^{2}+(x+1)^{2}}$.

Step 3: Evaluate at the crossing

\[ y'(-1) = \frac{-1}{(-1)^{2}+0^{2}} = -1. \]

Step 4: Convert the slope to an angle

The angle of inclination $\alpha \in [0^{\circ},180^{\circ})$ of a line of slope $k$ satisfies $\tan\alpha = k$. Here $k = -1$, and since $\arctan(-1) = -45^{\circ}$ lies outside that range, the inclination is

\[ \alpha = 180^{\circ} - 45^{\circ} = 135^{\circ} = \frac{3\pi}{4}. \]

The acute angle between the curve and the $x$-axis is therefore $45^{\circ}$.

Step 5: Verification

\[\text{Tangent line at } (-1,0):\quad y = -(x+1)\]

\[x = -0.99:\quad y = \arctan\!\left(1-\tfrac{1}{0.99}\right) \approx -0.010101; \qquad -(x+1) = -0.01 \;\;\checkmark\]