The problem consists of four questions related to Calculus. 1. a) Determine all the roots of three given equations. b) Given a function $g(x)$, find its inverse $g^{-1}(x)$, and show that $(g \circ g^{-1})(x) = x$.

1. Problem Description

The problem consists of four questions related to Calculus.

1. a) Determine all the roots of three given equations.

b) Given a function $g(x)$, find its inverse $g^{-1}(x)$, and show that $(g \circ g^{-1})(x) = x$.

2. a) Find the range and domain of two given functions.

b) Show two trigonometric identities.

3. a) Find all solutions to two trigonometric equations.

b) Determine whether two given functions are odd, even, or neither.

4. a) Find the derivative $\frac{dy}{dx}$ for two given functions.

b) Find the derivative $g'(x)$ of a given function $g(x)$.

c) Given $y = \arctan(\frac{1}{x})$, find $\frac{dy}{dx}$.

5. Solution Steps

1. a)

(i) $7 + 15e^{1-3x} = 10 \implies 15e^{1-3x} = 3 \implies e^{1-3x} = \frac{3}{15} = \frac{1}{5}$.

$1-3x = \ln(\frac{1}{5}) = -\ln(5) \implies 3x = 1 + \ln(5) \implies x = \frac{1 + \ln(5)}{3}$.

(ii) $f(x) = x^5 - x^4 - 32x^3 = 0 \implies x^3(x^2 - x - 32) = 0$. So $x=0$ is a triple root.

$x^2 - x - 32 = 0 \implies x = \frac{1 \pm \sqrt{1 + 4(32)}}{2} = \frac{1 \pm \sqrt{129}}{2}$.

So the roots are $x = 0$ (multiplicity 3), $x = \frac{1 + \sqrt{129}}{2}$, and $x = \frac{1 - \sqrt{129}}{2}$.

(iii) $x - xe^{3x+2} = 0 \implies x(1 - e^{3x+2}) = 0$. So $x=0$ is a root.

$1 - e^{3x+2} = 0 \implies e^{3x+2} = 1 \implies 3x+2 = \ln(1) = 0 \implies 3x = -2 \implies x = -\frac{2}{3}$.

The roots are $x = 0$ and $x = -\frac{2}{3}$.

b) $g(x) = \sqrt{x-3}$. To find the inverse, let $y = \sqrt{x-3}$. Then $y^2 = x-3 \implies x = y^2 + 3$.

So $g^{-1}(x) = x^2 + 3$, where $x \geq 0$ since the range of $g$ is $[0, \infty)$.

$(g \circ g^{-1})(x) = g(g^{-1}(x)) = g(x^2 + 3) = \sqrt{(x^2+3) - 3} = \sqrt{x^2} = |x| = x$ since $x \geq 0$.

2. a)

(i) $f(z) = \frac{|z| - 6}{|z| - 3}$.

For the domain, $|z| - 3 \neq 0 \implies |z| \neq 3 \implies z \neq 3$ and $z \neq -3$. Domain: $z \in \mathbb{R} \setminus \{-3, 3\}$.

If $|z| < 3$, $f(z) = \frac{|z|-6}{|z|-3} > 0$. If $|z| > 6$, $f(z) > 0$. If $3 < |z| < 6$, $f(z) < 0$.

$f(z) = \frac{|z|-3-3}{|z|-3} = 1 - \frac{3}{|z|-3}$. As $z \to \infty$, $f(z) \to 1$.

If $|z| = 0$, $f(z) = \frac{-6}{-3} = 2$. As $|z| \to 3^{-}$, $f(z) \to \infty$. As $|z| \to 3^{+}$, $f(z) \to -\infty$.

If $|z| = 6$, $f(z) = 0$. The range is $(-\infty, 0] \cup (1, \infty)$.

(ii) $f(z) = 2 + \sqrt{z^2+1}$. Since $z^2+1 \geq 1$, $\sqrt{z^2+1} \geq 1$. So $f(z) \geq 3$.

The domain is all real numbers. The range is $[3, \infty)$.

b)

(i) $1 + \tan^2\theta = \sec^2\theta$ is a standard trigonometric identity.

(ii) $\tan(\alpha + \beta) = \frac{\sin(\alpha + \beta)}{\cos(\alpha + \beta)} = \frac{\sin\alpha\cos\beta + \cos\alpha\sin\beta}{\cos\alpha\cos\beta - \sin\alpha\sin\beta}$.

Divide numerator and denominator by $\cos\alpha\cos\beta$:

$\tan(\alpha + \beta) = \frac{\frac{\sin\alpha\cos\beta}{\cos\alpha\cos\beta} + \frac{\cos\alpha\sin\beta}{\cos\alpha\cos\beta}}{\frac{\cos\alpha\cos\beta}{\cos\alpha\cos\beta} - \frac{\sin\alpha\sin\beta}{\cos\alpha\cos\beta}} = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$.

6. a)

(i) $\sin(2\theta) = \tan\theta \implies 2\sin\theta\cos\theta = \frac{\sin\theta}{\cos\theta} \implies 2\sin\theta\cos^2\theta = \sin\theta$.

$\sin\theta(2\cos^2\theta - 1) = 0 \implies \sin\theta\cos(2\theta) = 0$.

$\sin\theta = 0 \implies \theta = n\pi$, $n \in \mathbb{Z}$.

$\cos(2\theta) = 0 \implies 2\theta = \frac{\pi}{2} + n\pi \implies \theta = \frac{\pi}{4} + \frac{n\pi}{2} = \frac{(2n+1)\pi}{4}$, $n \in \mathbb{Z}$.

(ii) $4\sin^2(3t) - 3\sin(3t) = 1 \implies 4\sin^2(3t) - 3\sin(3t) - 1 = 0$.

$(4\sin(3t) + 1)(\sin(3t) - 1) = 0$.

$\sin(3t) = 1 \implies 3t = \frac{\pi}{2} + 2n\pi \implies t = \frac{\pi}{6} + \frac{2n\pi}{3}$, $n \in \mathbb{Z}$.

$\sin(3t) = -\frac{1}{4} \implies 3t = \arcsin(-\frac{1}{4}) + 2n\pi$ or $3t = \pi - \arcsin(-\frac{1}{4}) + 2n\pi$.

$t = \frac{1}{3}\arcsin(-\frac{1}{4}) + \frac{2n\pi}{3}$ or $t = \frac{\pi}{3} - \frac{1}{3}\arcsin(-\frac{1}{4}) + \frac{2n\pi}{3}$, $n \in \mathbb{Z}$.

b)

(i) $g(x) = -|x| + 2$. $g(-x) = -|-x| + 2 = -|x| + 2 = g(x)$. So $g(x)$ is even.

(ii) $f(x) = \frac{x^2}{x^4 - x} = \frac{x^2}{x(x^3 - 1)} = \frac{x}{x^3 - 1}$.

$f(-x) = \frac{-x}{(-x)^3 - 1} = \frac{-x}{-x^3 - 1} = \frac{x}{x^3 + 1}$.

Since $f(-x) \neq f(x)$ and $f(-x) \neq -f(x)$, $f(x)$ is neither even nor odd.

7. a)

(i) $y = \sin^2(4x) + \sec(x)$.

$\frac{dy}{dx} = 2\sin(4x)\cos(4x)(4) + \sec(x)\tan(x) = 8\sin(4x)\cos(4x) + \sec(x)\tan(x) = 4\sin(8x) + \sec(x)\tan(x)$.

(ii) $y = \cos(2x^2 - 1)$.

$\frac{dy}{dx} = -\sin(2x^2 - 1)(4x) = -4x\sin(2x^2 - 1)$.

b) $g(x) = (x+3)^2(2x^2+1)^3$.

$g'(x) = 2(x+3)(2x^2+1)^3 + (x+3)^2(3(2x^2+1)^2(4x)) = 2(x+3)(2x^2+1)^3 + 12x(x+3)^2(2x^2+1)^2$.

$g'(x) = 2(x+3)(2x^2+1)^2[(2x^2+1) + 6x(x+3)] = 2(x+3)(2x^2+1)^2[2x^2+1+6x^2+18x] = 2(x+3)(2x^2+1)^2[8x^2+18x+1]$.

c) $y = \arctan(\frac{1}{x})$.

$\frac{dy}{dx} = \frac{1}{1 + (\frac{1}{x})^2} \cdot (-\frac{1}{x^2}) = \frac{1}{1 + \frac{1}{x^2}} \cdot (-\frac{1}{x^2}) = \frac{1}{\frac{x^2+1}{x^2}} \cdot (-\frac{1}{x^2}) = \frac{x^2}{x^2+1} \cdot (-\frac{1}{x^2}) = -\frac{1}{x^2+1}$.

8. Final Answer

1. a) (i) $x = \frac{1 + \ln(5)}{3}$

(ii) $x = 0$ (multiplicity 3), $x = \frac{1 + \sqrt{129}}{2}$, $x = \frac{1 - \sqrt{129}}{2}$

(iii) $x = 0$, $x = -\frac{2}{3}$

b) $g^{-1}(x) = x^2 + 3$, $x \geq 0$. $(g \circ g^{-1})(x) = x$.

2. a) (i) Domain: $z \in \mathbb{R} \setminus \{-3, 3\}$. Range: $(-\infty, 0] \cup (1, \infty)$.

(ii) Domain: $z \in \mathbb{R}$. Range: $[3, \infty)$.

b) (i) $1 + \tan^2\theta = \sec^2\theta$ (standard identity)

(ii) $\tan(\alpha + \beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$

3. a) (i) $\theta = n\pi$, $\theta = \frac{(2n+1)\pi}{4}$, $n \in \mathbb{Z}$

(ii) $t = \frac{\pi}{6} + \frac{2n\pi}{3}$, $t = \frac{1}{3}\arcsin(-\frac{1}{4}) + \frac{2n\pi}{3}$, $t = \frac{\pi}{3} - \frac{1}{3}\arcsin(-\frac{1}{4}) + \frac{2n\pi}{3}$, $n \in \mathbb{Z}$

b) (i) Even

(ii) Neither

4. a) (i) $\frac{dy}{dx} = 4\sin(8x) + \sec(x)\tan(x)$

(ii) $\frac{dy}{dx} = -4x\sin(2x^2 - 1)$

b) $g'(x) = 2(x+3)(2x^2+1)^2[8x^2+18x+1]$

c) $\frac{dy}{dx} = -\frac{1}{x^2+1}$