Consider the function $$f(x) = \frac{e^x - 1}{e^x + 1}$$ (i) Show that $f(x)$ is increasing for all $x$ - HSC - SSCE Mathematics Extension 2 - Question 12 - 2017 - Paper 1

To show that $f(x)$ is an odd function, we need to prove that $f(-x) = -f(x)$:

1. Calculate $f(-x)$:

   $f(-x) = \frac{e^{-x} - 1}{e^{-x} + 1}$

2. Simplifying $f(-x)$ gives:

   $= \frac{\frac{1}{e^x} - 1}{\frac{1}{e^x} + 1} = \frac{1 - e^x}{1 + e^x}$

3. This can be rewritten as:

   $= -\frac{e^x - 1}{e^x + 1} = -f(x)$

Thus, $f(-x) = -f(x)$, confirming that $f(x)$ is an odd function.

As $x$ approaches infinity, the exponential terms dominate:

$f(x) = \frac{e^x - 1}{e^x + 1} \approx \frac{e^x}{e^x} = 1$

Thus, as $x \to \infty$, $f(x) \to 1$. This indicates that $f(x)$ approaches 1 for large positive values of $x$.

The graph of $f(x)$ consists of:

• An increasing function for all $x$.
• Intersects the y-axis at $f(0) = 0$.
• Approaches 1 as $x \to \infty$.

Therefore, the graph should be a smooth curve starting at the origin, increasing and approaching the horizontal line $y=1$.

For $y = \frac{1}{f(x)}$, we note:

• Since $f(x)$ approaches 1, $y$ approaches 1 as well.
• As $f(x)$ is positive and increasing, $y = \frac{1}{f(x)}$ is a decreasing function.

The graph will start from $y = 1$ and approach infinity as $x \to 0^+$ and will decrease towards 1 as $x \to \infty$.

To solve the quadratic equation, we employ the quadratic formula:

$z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = 2 + 3i$, and $c = 1 + 3i$.

Calculate the discriminant:

$b^2 - 4ac = (2 + 3i)^2 - 4(1)(1 + 3i) = (4 + 12i - 9) - (4 + 12i) = -5$

Thus,

$z = \frac{-(2 + 3i) \pm \sqrt{-5}}{2} = \frac{-2 - 3i \pm i\sqrt{5}}{2}$

Giving us the solutions: $z = -1 - \frac{3}{2} i \pm \frac{\sqrt{5}}{2} i$

To solve the integral, we use integration by parts. Let:

• $u = \tan^{-1}(x)$, then $du = \frac{1}{1+x^2} dx$.
• $dv = x \, dx$, then $v = \frac{x^2}{2}$.

Applying integration by parts:

$\int u \, dv = uv - \int v \, du$

This gives:

d$= \tan^{-1}(x) \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot \frac{1}{1+x^2} \, dx$

Resimplifying:

$= \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \int \frac{x^2}{1+x^2} \, dx$

Decomposing the integrand yields:

$= \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \int (1 - \frac{1}{1+x^2}) \, dx$

Evaluating the integral:

$= \frac{x^2}{2} \tan^{-1}(x) - \frac{1}{2} \left(x - \tan^{-1}(x) \right) + C$

If $(x - a^2)$ is a factor of $P(x)$, then by the factor theorem:

$P(a^2) = 0$

Additionally, the function is polynomial, and thus by substituting $x = a$:

$P(a)$

To show that also equals zero, we substitute and confirm through polynomial division or evaluation at the roots,$P(a) = 0$ as well. This establishes $a^2$ as a root.

Given that $P(a^2) = 0$, we have: $P(a^2) = (a^2)^4 - 3(a^2)^3 + (a^2)^2 + 4 = 0$

Substituting $y = a^2$ simplifies to: $y^4 - 3y^3 + y^2 + 4 = 0$

To find the value of $y$, we can use numerical or graphical methods (if necessary) or apply the rational root theorem for integer possibilities.

Testing roots, we find:

• Trial with $y = 2$ or direct evaluation may lead us to confirm valid integers that satisfy setting $y$.

This will lead us to $a = \sqrt{2}$ or suitable candidates based on polynomial factorizations or numeric validations.