Sketch the curve with equation $y = \ln(x+1), x > -1$, stating the equation of asymptote and intercepts with the axes. On the same diagram, sketch the curve with equation $y = \sqrt{9-x^2}, -3 \le x \le 3$, stating the intercepts with the axes. Use your answer in part (i), solve the inequality $\ln(x+1) < \sqrt{9-x^2}$. Hence solve the inequality $\ln(x^2+1) < \sqrt{9-x^4}$.

1. Identify the asymptote and intercepts of $y = \ln(x+1)$.

For $y = \ln(x+1)$:
The domain is $x > -1$. As $x \to -1^+$, $y \to -\infty$.
The vertical asymptote is $x = -1$.
To find intercepts:
When $x = 0$, $y = \ln(1) = 0$.
When $y = 0$, $\ln(x+1) = 0 \implies x+1 = 1 \implies x = 0$.
The curve passes through the origin $(0,0)$.

2. Identify the shape and intercepts of $y = \sqrt{9-x^2}$.

For $y = \sqrt{9-x^2}$ over $-3 \le x \le 3$:
Squaring both sides gives $y^2 = 9 - x^2 \implies x^2 + y^2 = 9$.
Since $y \ge 0$, this represents the upper half of a circle centred at $(0,0)$ with radius $3$.
To find $x$-intercepts: set $y = 0 \implies x^2 = 9 \implies x = \pm 3$. The intercepts are $(-3,0)$ and $(3,0)$.
To find the $y$-intercept: set $x = 0 \implies y = \sqrt{9} = 3$. The intercept is $(0,3)$.

3. Sketch both graphs on the same coordinate axes.

Draw the $x$ and $y$ axes.
Draw and label the vertical dotted line $x = -1$.
Draw the logarithmic curve starting from the right of $x = -1$, passing through $(0,0)$, and increasing with a decreasing gradient.
Draw the upper semicircle from $(-3,0)$ through $(0,3)$ to $(3,0)$.
Ensure the two curves intersect once in the first quadrant (between $x=0$ and $x=3$) and clearly label the equations of the curves and all stated intercepts.

1. Relate the inequality to the relative positions of the graphs

The inequality $\ln(x+1) < \sqrt{9-x^2}$ is satisfied for the range of $x$ where the curve $y = \ln(x+1)$ lies strictly below the curve $y = \sqrt{9-x^2}$.

2. Find the point of intersection of the two curves

Using a graphing calculator, we solve $\ln(x+1) = \sqrt{9-x^2}$.
The $x$-coordinate of the point of intersection in the valid domains is $x \approx 2.6997$.

3. Determine the solution interval using the domain of the functions

From part (i), the domain of $y = \ln(x+1)$ is $x > -1$ (vertical asymptote at $x = -1$). From the graphs, the logarithmic curve starts from $-\infty$ and lies below the semi-circle up until the intersection point.
$\therefore -1 < x < 2.70$ (to 3 s.f.)

1. Determine the solution for part (ii) to use in part (iii).

Using a graphing calculator, the intersection of $y = \ln(x+1)$ and $y = \sqrt{9-x^2}$ for $x > -1$ is $x \approx 2.6997$.
From the graphs in part (i), $y = \ln(x+1)$ lies below $y = \sqrt{9-x^2}$ for the interval between the vertical asymptote and the intersection.
Hence, the solution to $\ln(x+1) < \sqrt{9-x^2}$ is $-1 < x < 2.6997$.

2. Relate the inequality in part (iii) to the result of part (ii).

The inequality $\ln(x^2+1) < \sqrt{9-x^4}$ can be written as $\ln(x^2+1) < \sqrt{9-(x^2)^2}$.
This is identical to the inequality in (ii) with $x$ substituted by $x^2$.
Substituting $x^2$ into the solution range from (ii), we require:
$-1 < x^2 < 2.6997$

3. Solve for $x$ to find the final range.

Since $x^2 \ge 0$ for all real $x$, the condition $x^2 > -1$ is always true.
We therefore only need to solve $x^2 < 2.6997$:
$-\sqrt{2.6997} < x < \sqrt{2.6997}$
$-1.6431 < x < 1.6431$
So, the solution is $-1.64 < x < 1.64$ (to 3 s.f.).