A Collection of Problems in Differential Calculus: Problems from Calculus I Final Examinations, 2000-2020 Department of Mathematics, Simon Fraser University

Let \(f(x)=\ln x-cx^2\text{.}\) Note that the domain of \(f\) is the interval \((0,\infty )\) and that \(\ds \lim _{x\to \pm \infty }f(x)=-\infty\text{.}\) From \(\ds f'(x)=\frac{1}{x}-2cx\) it follows that there is a local (and absolute) maximum at \(\ds x=\frac{1}{\sqrt{2c}}\text{.}\) Since the function \(f\) is continuous on its domain, by the Intermediate Value Theorem, it will have a root if and only if \(\ds f\left( \frac{1}{\sqrt{2c}}\right) >0\text{.}\) Thus

Note that the function \(y=3x^3+2x+12\text{,}\) as a polynomial, is continuous on the set of real numbers. Also, \(\ds \lim _{x\to -\infty }(3x^3+2x+12)=-\infty\) and \(\ds \lim _{x\to \infty }(3x^3+2x+12)=\infty\text{.}\) By the Intermediate Value Theorem, the function has at least one root. Next, note that \(y'=6x^2+2>0\) for all \(x\in \mathbb{R}\text{.}\) This means that the function is increasing on its domain and therefore has only one root. (If there is another root, by Rolle's theorem there will be a critical number.)

Let \(y=x^3+9x+5\text{.}\) This function is a polynomial, and so is continuous. Since

by the Intermediate Value Theorem that \(y \) has at least one root in the interval \((-1,0) \text{.}\) Furthermore, since

we see that \(y \) is increasing for all \(x \in \mathbb{R} \text{.}\) Therefore, \(y\) can have no additional roots. Hence, \(x^3+9x+5=0\) has exactly one real solution.

Note that the function \(f\) is continuous on its domain \(\mathbb{R}\backslash\{-1\}\text{.}\) Since \(\ds \lim _{x\to \infty }\frac{1}{(x+1)^2}=0\) and \(|\sin x|\leq 1\text{,}\) for all \(x\in \mathbb{R}\text{,}\) it follows that \(\ds \lim _{x\to \infty }f(x)=\lim _{x\to \infty }(-2x)=- \infty\text{.}\) Also, \(\ds f(0)=1>0\text{.}\) By the Intermediate Value Theorem, the function has at least one root in the interval \((0,\infty )\text{.}\) Next, note that \(\ds f'(x)=-\frac{2}{(x+1)^3} -2+\cos x\lt 0\) for all \(x\in (0,\infty)\text{.}\) This means that the function is decreasing on \((0,\infty)\) and therefore has only one root.

1. The domain of \(f(x)=\arcsin x\) is the interval \([-1,1]\) and its range is \(\ds \left[ -\frac{\pi }{2},\frac{\pi }{2}\right]\text{.}\)

2. For \(x\in (-1,1)\) let \(y=\arcsin x\text{.}\) Then \(\sin y=x\) and \(y'\cos y=1\text{.}\) Since \(\ds y \in \left( -\frac{\pi }{2},\frac{\pi }{2}\right)\) it follows that \(\cos y>0\text{.}\) Thus \(\ds \cos y =\sqrt{1-\sin ^2y}=\sqrt{1-x^2}\text{.}\)

3. The domain of the function \(g\) is the set of all real numbers and its range is the set \(\ds \left[ -\frac{\pi }{2},\frac{\pi }{2}\right]\text{.}\)

   

4. \(\ds g'(x)=\left\{ \begin{array}{rl} 1& \mbox{if } x\in (4n-1,4n+1), \ n\in \mathbb{Z}\\ -1& \mbox{if } x\in (4n+1,4n+3), \ n\in \mathbb{Z}\\ \end{array} \right.\text{.}\) The derivative of \(g\) is not defined if \(x=2n+1\text{,}\) \(n\in \mathbb{Z}\text{.}\)

5. The function \(F(x)=4x-2+\cos \left( \frac{\pi x}{2}\right)\) is continuous on the set of real numbers. From \(\ds \lim _{x\to \pm \infty }F(x)=\pm \infty\text{,}\) by the Intermediate Value Theorem, the function \(F\) has at least one root. From \(\ds F'(x)=1-\frac{2}{\pi }\sin \left( \frac{\pi x}{2}\right)>0\) we conclude that \(F\) is monotone.

1. We have

   \begin{equation*} \tanh x= \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x+e^{-x}} = \frac{1-e^{-2x}}{1+e^{-2x}}. \end{equation*}

   Hence,

   \begin{equation*} \lim_{x\to\infty}\tanh = \lim_{x\to\infty} \frac{1-e^{-2x}}{1+e^{-2x}} = 1. \end{equation*}

2. We differentiate:

   \begin{equation*} \frac{dy}{dx}\big\vert_{x=0} = \left[\frac{d}{dx}\cosh x \right]_{x=0} +3 = 3. \end{equation*}

   Therefore, the equation of the tangent line to \(y=\cosh(x)+3x+4\) at the point \((0,5)\) is given by

   \begin{equation*} y_T = 3 x + 5. \end{equation*}

1. Take \(y=-x\) for all \(x \in \mathbb{R}.\) Then \(y'(x) = -1 \lt 0 \) for all \(x \in \mathbb{R}\text{,}\) and so the function is strictly decreasing on its entire domain.

2. Take \(y=\frac{1}{x^2}\) for \(x \in [-1, 1] \text{.}\) Then there is no number \(c\in (-1,1)\) such that \(f'(c) = \frac{1-1}{-2} = 0\text{.}\) The MVT fails because the function is not continuous on \([-1,1]\text{.}\)

3. Let \(f(x) = \sqrt{x}\text{.}\) Then we know that \(f(100)=10\) and \(f'(100)=1/20.\) Therefore, we compute:

   \begin{equation*} \sqrt{102}=f(102) \approx f(100) + f'(100)(2) = 10 + \frac{1}{10} =10.1. \end{equation*}

   Since \(f(x) \) is concave down, this will be an overestimate. Indeed, using computer software, we find that \(\sqrt{102} = 10.0995...\text{.}\)

For all \(x \in \mathbb{R},\) we have that

Hence, \(\sinh(x)\) is an odd function.