The diagram shows the graph of $y = f(x)$ - HSC - SSCE Mathematics Extension 2 - Question 3 - 2003 - Paper 1

To sketch the graph of $y = f(x) + f'(x)$, remember:

1. The graph $f'(x)$ represents the slope of $f(x)$, indicating where the function is increasing or decreasing.
2. At points where $f'(x) = 0$, the graph could have local maxima or minima, and these points should be included in the sketch.
3. The combined graph will have features of both $f(x)$ and its derivative. Analyze crucial points, such as the zeros of $f'(x)$ and the local extrema of $f(x)$, to accurately depict this.

For the graph of $y = e^{f(x)}$:

1. This transformation indicates an exponential function, which is always positive regardless of $f(x)$.
2. Identify critical points from $f(x)$ to understand the behavior of $e^{f(x)}$. The values of $f(x)$ will greatly influence the steepness and direction of the exponential curve.
3. If $f(x)$ has any x-intercepts, those will become points where $e^{f(x)} = 1$.
4. Sketch these transformations comprehensively.

Given the ellipse equation, $\frac{x^2}{9} + \frac{y^2}{4} = 1$:

1. Identify the semi-major axis $a = 3$ and semi-minor axis $b = 2$ from the denominators.
2. Calculate the eccentricity $e$ using the formula: $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{9}} = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}.$
3. The foci are located at $(\pm ae, 0) = \left(\pm 3 \cdot \frac{\sqrt{5}}{3}, 0\right) = (\pm \sqrt{5}, 0).$
4. The equations of the directrices are given by $x = \pm \frac{a}{e} = \pm \frac{3}{\sqrt{5}}.$

To find the area of the annulus:

1. Determine the bounds of the curve $y = (x-1)(3-x)$ for intersection with the x-axis.
2. Integrate to find the radii of revolution about the line $x = 3$:
   • Outer radius: $R_o = 3 - x_{left}$
   • Inner radius: $R_i = 3 - x_{right}$
3. Calculate the area of the annulus: $Area = \pi (R_o^2 - R_i^2).$
4. After simplifications, confirm that the area simplifies to $4\pi\sqrt{1-y}$.

To determine the volume:

1. Use the previously found area function for the annulus, $A(y) = 4\pi \sqrt{1 - y}$.
2. Integrate this area function over its bounds, noting the limits defined by $y$ in the interval where the region exists.
3. Set up the integral: $V = \int A(y) \, dy = \int (4\pi \sqrt{1 - y}) \, dy.$
4. Solve the integral: After evaluating, this will yield the final volume of the solid formed by rotation.