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 $f'(x)$ to zero:

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

This simplifies to:

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

Multiplying both sides by $e^{2x}$ gives:

$e^{x} = 4$

Thus, we have:

$x = ext{ln}(4)$.

To find the y-coordinate at the maximum turning point:

$f( ext{ln}(4)) = e^{- ext{ln}(4)} - 2e^{-2 ext{ln}(4)}$

This results in:

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

Thus, the coordinates of the maximum turning point are ( ext{ln}(4), rac{1}{8}).

To evaluate $f(2)$:

$f(2) = e^{-2} - 2e^{-4}$

Calculating this gives:

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

As $x \to -\infty$, both exponential terms $e^{-x}$ and $e^{-2x}$ approach infinity. Therefore, the function $f(x)$ tends to:

$f(x) \to \infty.$

The y-intercept occurs when $x = 0$:

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

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

The graph should show:

• A maximum turning point at ( ext{ln}(4), rac{1}{8}).
• The y-intercept at $(0, -1)$.
• Behaviour tending to infinity as $x$ approaches negative infinity.

By the Inscribed Angle Theorem, we know that the angle at the center of the circle is twice that of the angle at the circumference subtended by the same arc. Thus:

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

To prove that quadrilateral $ACDO$ is cyclic, we need to show that opposite angles are supplementary. Since $\angle AOC = 2x$ and $\angle ADC = x$, we see that:

$\angle AOC + \angle ADC = 2x + x = 3x,$
which does not tell us supplementary. Therefore, we check the relationship with respect to diameter, since $D$ lies on $BC$, it suggests that angles $ACD$ and $AOD$ should be supplementary.

For collinearity, the slopes of both segments $PM$ and $MO$ must be equal. Given that $M$ is the midpoint of $AC$, and $P$ is the circle's center, the coordinates of points can be used to show the linear relationship, confirming the alignment of points $P, M, O$.