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Calculus 1: midterm 1 solutions

This is for a test for Calculus 1: course.

Question 1

(a) Represent function $h(x) = \sqrt{x^{2} - 1}$ as a composition of two functions $f$ and $g$. (b) Provide the formula for the composition $f(g(x))$ of $f(x) = x^{2} + x$ and $g(x) = 2x-1$.

$$f(y) = \sqrt{y}, \qquad g(x) = x^{2} - 1$$ then $$h(x) = f(g(x))$$

$$f(g(x)) = (2x + 1)^{2} + (2x-1)$$

Question 2

Sketch the graphs of three functions with the three main types of discontinuties. Describe these discontinuities with limits.

Question 3

(a) State the $\epsilon - \delta$ definition of a limit. (b) Use the defintion to prove that $\lim\limits_{x \to 0} (\sqrt[3]{x}) = 0$

For any $\epsilon > 0$ there is $\delta$ such that $$0 < |x-a | < \delta \Rightarrow |f(x) - f(a)| < \epsilon$$

We need to prove that given $\epsilon > 0$, we can find $\delta$ such that $$0 < |x - 0 | < \delta \Rightarrow |\sqrt[3]{x} - \sqrt[3]{0}| < \epsilon$$ or $$0 < |x | < \delta \Rightarrow |\sqrt[3]{x}| < \epsilon$$ or $$0 < |x | < \delta \Rightarrow |x| < \epsilon^{3}$$ Choose $\delta = \epsilon^{3}$. Retrace the steps to finish the proof.

Question 4

Find the vertical asymptotes of the function $$f(x) = \frac{x - 1}{(x^{2} -1)(x+2)}$$ by computing necessary limits.

\begin{aligned} f(x) & = \frac{x - 1}{(x^{2} -1)(x+2)} \\ &= \frac{x - 1}{(x -1)(x+1)(x+2)} \end{aligned} The denominator is 0 for $x = 1, -1, -2$. These are the conditions for the vertical asymptotes. $$\lim_{x \to 1} \frac{x - 1}{(x - 1)(x + 1)(x + 2)} = \underbrace{\lim_{x \to 1} \frac{1}{(x + 1)(x + 2)}}_{\text{Continuous at 1}} = \frac{1}{(1 + 1)(1 + 2)} =\frac{1}{6}$$ So, $x = \frac{1}{6}$ is not an asymptote. $$\lim_{x \to -1} \frac{x - 1}{(x - 1)(x + 1)(x + 2)} = \infty$$ as $\text{numerator } \to -2$ and $\text{denominator } \to 0$. So, $x = -1$ is a vertical asymptote. $$\lim_{x \to -2} \frac{x - 1}{(x - 1)(x + 1)(x + 2)} = \infty$$ as $\text{numerator } \to -3$ and $\text{denominator } \to 0$. So, $x = -2$ is a vertical asymptote.

Question 5

Evaluate the limit $$\lim_{x \to \infty} \frac{x^{3} - 1}{2x^{3} -x + 5}$$ What is its graphical meaning?

\begin{aligned} \lim_{x \to \infty} \frac{x^{3} - 1}{2x^{3} -x + 5} & = \lim_{x \to \infty} \frac{\frac{x^{3} - 1}{x^{3}}}{\frac{2x^{3} - x + 5}{x^{3}}} \\ &= \frac{1 - \frac{1}{x^{3}}}{2 - \frac{1}{x^{2}} + \frac{5}{x^{3}}} \\ &= \frac{1 - 0}{2 - 0 + 0} = \frac{1}{2} \end{aligned} So, $y = \frac{1}{2}$ is a horizontal asymptote.

Question 6

The graph of a function $f$ is given below. Sketch the graph of $$g(x) = 2f(x + 2) + 3$$ Explain how you get it.

$$g(x) = \underset{(2)}{2} f( \underset{(1)}{x + 2} ) + \underset{(3)}{3}$$

(1)
Shift left 2 units
(2)
Stretch vertical by 2
(3)
Shift up by 3 points

Question 7

The pictured graph represents the altitude (in thousands of feet) of a plane above the ground at time $x$ (in sec). (a) What does the slope of the curve represent? (b) Describe what happened to the plane. Explain.

The slope at each point represents the speed of descent (ascent) of the plane.

First the plane descends fast, then slows down and lands safely.

Question 8

From the definition, compute the derivative of $f(x) = x^{2} + 1$ at $a = 2$.

\begin{aligned} f^{\prime}(2) &= \lim_{h \to 0} \frac{f(2 + h) - f(2)}{h} \\ &= \lim_{h \to 0} \frac{\left[ (2 + h)^{2} + 1 \right] - \left[ 2^{2} + 1 \right]}{} \\ &= \lim_{h \to 0} \frac{4 + 4h + h^{2} + 1 - 4 -1}{h} \\ &= \lim_{h \to 0} \frac{4h + h^{2}}{h} \\ &= \lim_{h \to 0} (4 + h) = 4 \end{aligned}
The graph of $f(x)$ is given below. Estimate the values of the derivative $f^{\prime}(x)$ for $x = 0,4,6$.
Draw the tangent lines, then compute their slopes as $\frac{\text{rise}}{\text{run}}$, using the grid.
\begin{aligned} f^{\prime}(0)&\approx \frac{6}{2}=3 \\ f^{\prime}(4)&\approx \frac{-9}{9}=-1 \\ f^{\prime}(6)&\approx \frac{-6}{2}=-3 \end{aligned}