Consider the function $f(x) = e^{-x} - 2e^{-2x}.$ (i) Find $f'(x)$ - HSC - SSCE Mathematics Extension 1 - Question 4 - 2011 - Paper 1

To find the maximum turning point, we set the derivative equal to zero:

$-e^{-x} + 4e^{-2x} = 0$

Solving this gives:

$e^{-x} = 4e^{-2x} \Rightarrow 4 = e^{-x + 2x} \Rightarrow 4 = e^{x}\Rightarrow x = ext{ln}(4)$

Now, substituting into the original function to find the y-coordinate:

$f( ext{ln}(4)) = e^{- ext{ln}(4)} - 2e^{-2 ext{ln}(4)} = \frac{1}{4} - 2 \cdot \frac{1}{16} = \frac{1}{4} - \frac{1}{8} = \frac{1}{8}$

Thus, the coordinates of the maximum turning point are $(\text{ln}(4), \frac{1}{8})$.

We evaluate the function at $x = 2$:

$f(2) = e^{-2} - 2e^{-4} = \frac{1}{e^2} - 2 \cdot \frac{1}{e^4} = \frac{1}{e^2} - \frac{2}{e^4} = \frac{e^2 - 2}{e^4}$

As x o - hinspace orall, we have:

f(x) = e^{-x} - 2e^{-2x} \to hinspace orall - 0 = hinspace orall

This means the function approaches infinity.

To find the y-intercept, we evaluate:

$f(0) = e^{0} - 2e^{0} = 1 - 2 = -1$

Thus, the y-intercept is at $(0, -1)$.

In sketching the graph, indicate the maximum turning point at $(\text{ln}(4), \frac{1}{8})$ and the y-intercept at $(0, -1)$. The graph should show an increasing trend towards the maximum and then decrease towards negative infinity as $x$ approaches infinity.

Since $D$ lies on $BC$, it follows from the Inscribed Angle Theorem that:

$\angle AOC = 2 \angle ABC = 2x$.

For $ACDO$ to be a cyclic quadrilateral, we need to show that opposite angles sum to $180^{\circ}$. Since:

$\angle AOC + \angle ADB = 2x + x = 3x$

If $x = 60^{\circ}$, then this holds true.

To show collinearity, we note that since $M$ is the midpoint of $AC$ and $P$ is the circumcenter, all points lie on the line connecting $O$ and the midpoint of line segment $AC$, establishing collinearity.