Name: Solution GuideMath 241: Calculus & Analytic Geometry A
Exam 1
4 October 2002

Always the beautiful answer who asks a more beautiful question.
-E. E. Cummings

Instructions: Show all work to receive full or partial credit. All University rules and guidelines for student conduct are applicable.

1.

[20 pts] Find the equation of the hyperbola that passes through the points (-1,0) and (1,0), and has asymptotes of y=2x and y=-2x, as shown.

The standard form for a hyperbola with this orientation is

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,$$

and we simply need to find a and b. Since it passes through (-1,0) and (1,0), we see that a=1. The slopes of the asymptotes will be $\pm \frac{b}{a}$. Therefore, we need b=2 to match the diagram, so that the equation is

$$x^2 - \frac{y^2}{4} = 1,$$

2.

[20 pts] For what value of c is f(x) continuous on the whole real line? Why?

$$f(x) = \begin{cases} \frac{\sin(2x)}{x} & x<0 \\ x + c & x \geq 0 \end{cases}$$

For a function to be continuous at a point x=a:

• f(a) must exist.
• $\lim_{x \rightarrow a} f(x)$ must exist. This means you must check the limit from each side in cases like x=0.
• The limit and the value of the function must be the same.

We see that the function is defined to be the ratio of a sine function and a polynomial on the left. Since there are no singularities on the left because x=0 is not in the domain, the function is continuous on $(-\infty,0)$. Similarly, polynomials are continuous so the f(x) is also continuous on $(0,\infty)$. We just need to see what happens at x=0 where the two domains meet.

We see that

f(0) = c,

and that

$$\lim_{x\rightarrow0^+} f(x) = c.$$

On the left, we find that

$$\lim_{x\rightarrow0^-} \frac{\sin(2x)}{x} = \lim_{x\rightarro... ... 2 \cos x \cdot \lim_{x\rightarrow0^-} \frac{\sin(x)}{x} = 2.$$

Note: You can only break up products like this when you know that both limits exist.

Thus, for f(x) to be continuous, c=2.

3.

[20 pts] Determine an equation for the line tangent to the curve

$$f(x) = x^2 + \sin x$$

at $x=\frac{\pi}{3}$. Simplify your solution as much as possible.

Anticipating that we are going to use point-slope form for the tangent line, we need a point on the curve. Only $x=\frac{\pi}{3}$ is given, we must find

$$f\left(\frac{\pi}{3}\right) = \frac{\pi^2}{9} + \sin\left(\fr... ...^2}{9} + \frac{\sqrt{3}}{2} = \frac{2 \pi^2 + 9 \sqrt{3}}{18}.$$

Now that we have a point on the curve, we must find its slope $m=f'\left(\frac{\pi}{3}\right)$.
$$\begin{gather}f'(x) = 2x + \cos(x) \\ m = \frac{2 \pi}{3} + \frac{1}{2} = \frac{4 \pi + 3}{6} \end{gather}$$

Now, we have all the information that we need. An equation for the tangent line would be

$$\left(y - \frac{2 \pi^2 + 9 \sqrt{3}}{18}\right) = \frac{4 \pi + 3}{6} \left(x- \frac{\pi}{3} \right).$$

4.

[10 pts] Evaluate the following limit:

$$\lim_{x \rightarrow 2} \frac{(x^2+2x-8)^2}{x^2-4}.$$

$$\begin{eqnarray*}\lim_{x \rightarrow 2} \frac{(x^2+2x-8)^2}{x^2-4} & = & \lim_{... ... \lim_{x \rightarrow 2} \frac{(x+4)^2 (x-2)}{(x+2)} \\ & = & 0. \end{eqnarray*}$$

5.

[10 pts] Evaluate the following limit:

$$\lim_{x \rightarrow 0} \frac{2 \sin x-\sin(2 x)}{x^3}.$$

$$\begin{eqnarray*}\lim_{x \rightarrow 0} \frac{2 \sin x-\sin(2 x)}{x^3} & = & \li... ...frac{1-\cos x }{x^2} \\ & = & 2 \cdot \frac{1}{2} \\ & = & 1. \end{eqnarray*}$$

Once again, we can only break up limits like this, when we know that the individual limits exist.

6.

[10 pts] Evaluate the derivative of f(x) where

$$f(x) = \frac{\tan(x)}{x^2 - 1/x}.$$

Here, we apply the quotient rule.

$$f'(x) = \frac{\sec^2(x) (x^2 - 1/x) - \tan (x) (2x+x^{-2})}{(x^2 - 1/x)^2}.$$

7.

[10 pts] Evaluate the derivative of f(x) where

$$f(x) = (x^2 - 1/x)\cos(x).$$

Here we apply the product rule.

$$f'(x) = (2x+x^{-2})\cos x- (x^2 - 1/x)\sin x.$$