Math 161 Fall 2014 Review Guide I
Introduction
This review guide covers the fundamental concepts from the first half of Math 161 (Calculus I) for Fall 2014. It includes limits, derivatives, and their applications. Use this guide to refresh your understanding before exams.
1. Limits
1.1 Definition
The limit of a function f(x) as x approaches a, denoted as lim(xa) f(x), is the value that f(x) approaches as x gets arbitrarily close to a.
1.2 Properties of Limits
| Property | Rule |
|---|---|
| Limit of a Sum | lim(xa) [f(x) + g(x)] = lim(xa) f(x) + lim(xa) g(x) |
| Limit of a Product | lim(xa) [f(x) g(x)] = lim(xa) f(x) lim(xa) g(x) |
| Limit of a Quotient | lim(xa) [f(x)/g(x)] = lim(xa) f(x)/lim(xa) g(x), provided lim(xa) g(x) 0 |
| Limit of a Power | lim(xa) [f(x)]^n = [lim(xa) f(x)]^n, for n a positive integer |
1.3 Special Limits
lim(x0) (1 + x)^(1/x) = e
lim(x) (1 + 1/x)^x = e
1.4 Techniques for Evaluating Limits
Example 1: Direct Substitution
Evaluate lim(x2) (3x - 4x + 1)
Solution: Direct substitution gives 3(2) - 4(2) + 1 = 12 - 8 + 1 = 5
Example 2: Factoring
Evaluate lim(x3) (x - 9)/(x - 3)
Solution: Factor the numerator to get lim(x3) (x - 3)(x + 3)/(x - 3) = lim(x3) (x + 3) = 6
Example 3: Rationalizing
Evaluate lim(x0) ((x+1) - 1)/x
Solution: Multiply by the conjugate: lim(x0) (((x+1) - 1)((x+1) + 1))/[x((x+1) + 1)] = lim(x0) (x+1-1)/[x((x+1) + 1)] = lim(x0) 1/((x+1) + 1) = 1/2
2. Continuity
2.1 Definition
A function f is continuous at a point a if:
- f(a) is defined
- lim(xa) f(x) exists
- lim(xa) f(x) = f(a)
2.2 Types of Discontinuities
- Removable (hole): The limit exists but is not equal to the function value
- Jump: Left and right limits exist but are not equal
- Infinite: The function approaches infinity at some point
- Essential: No limit exists at the point
Example: Finding Discontinuities
Find the discontinuities of f(x) = (x - 4)/(x - 2)
Solution: f(x) = (x - 2)(x + 2)/(x - 2) = x + 2 for x 2. This function has a removable discontinuity at x = 2 because the limit exists (4) but the function is not defined there.
3. Derivatives
3.1 Definition
3.2 Differentiation Rules
| Rule | Formula |
|---|---|
| Power Rule | d/dx(x^n) = nx^(n-1) |
| Constant Rule | d/dx(c) = 0 |
| Constant Multiple | d/dx[cf(x)] = cf'(x) |
| Sum Rule | d/dx[f(x) + g(x)] = f'(x) + g'(x) |
| Product Rule | d/dx[f(x)g(x)] = f'(x)g(x) + f(x)g'(x) |
| Quotient Rule | d/dx[f(x)/g(x)] = [f'(x)g(x) - f(x)g'(x)]/[g(x)] |
| Chain Rule | d/dx[f(g(x))] = f'(g(x))g'(x) |
3.3 Derivatives of Basic Functions
| Function | Derivative |
|---|---|
| sin(x) | cos(x) |
| cos(x) | -sin(x) |
| tan(x) | sec(x) |
| e^x | e^x |
| ln(x) | 1/x |
| a^x (a > 0, a 1) | a^x ln(a) |
3.4 Examples
Example 1: Power Rule
Find f'(x) for f(x) = 3x - 2x + 5x - 7
Solution: f'(x) = 12x - 6x + 5
Example 2: Product Rule
Find f'(x) for f(x) = x sin(x)
Solution: f'(x) = (x)' sin(x) + x (sin(x))' = 2x sin(x) + x cos(x)
4. Implicit Differentiation
Implicit differentiation is used when a function is not given in the form y = f(x) but instead is given implicitly, such as x + y = 25.
Procedure
- Differentiate both sides of the equation with respect to x
- Whenever you differentiate a y term, multiply by dy/dx
- Solve for dy/dx
Example
Find dy/dx for the equation x + y = 25
Solution: Differentiating both sides: 2x + 2y(dy/dx) = 0
Solving for dy/dx: dy/dx = -x/y
5. Related Rates
Related rates problems involve finding the rate at which one quantity changes given the rate at which another related quantity changes.
Procedure
- Identify what quantities are changing and how they are related
- Write an equation relating the quantities
- Differentiate both sides with respect to time
- Substitute known values and solve for the unknown rate
Example
A ladder 10 feet long rests against a wall. If the bottom of the ladder slides away from the wall at a rate of 1 ft/s, how fast is the top of the ladder sliding down the wall when the bottom is 6 feet from the wall?
Solution: Let x be the distance from the wall to the bottom of the ladder and y be the height of the top of the ladder. By the Pythagorean theorem, x + y = 100.
Differentiating: 2x(dx/dt) + 2y(dy/dt) = 0
We know dx/dt = 1 ft/s and x = 6 ft. From the Pythagorean theorem, y = (100 - 36) = 8 ft.
Substituting: 2(6)(1) + 2(8)(dy/dt) = 0, so dy/dt = -12/16 = -3/4 ft/s (negative because y is decreasing)
6. Applications of Derivatives
6.1 Tangent Lines
The equation of the tangent line to the graph of y = f(x) at x = a is:
6.2 Increasing and Decreasing Functions
- If f'(x) > 0 on an interval, then f is increasing on that interval
- If f'(x) < 0 on an interval, then f is decreasing on that interval
6.3 Local Extrema
A function has a local maximum at c if f(c) f(x) for all x in some open interval containing c.
A function has a local minimum at c if f(c) f(x) for all x in some open interval containing c.
First Derivative Test for Local Extrema
Suppose f is continuous at c and differentiable on an open interval containing c (except possibly at c). If f'(x) changes from positive to negative at c, then f has a local maximum at c. If f'(x) changes from negative to positive at c, then f has a local minimum at c.
Second Derivative Test for Local Extrema
Suppose f'' is continuous near c. If f'(c) = 0 and f''(c) > 0, then f has a local minimum at c. If f'(c) = 0 and f''(c) < 0, then f has a local maximum at c.
6.4 Optimization Problems
Optimization problems involve finding maximum or minimum values of a function.
Example
Find two numbers whose product is 100 and whose sum is a minimum.
Solution: Let x and y be the numbers. We know xy = 100, so y = 100/x. The sum is S = x + y = x + 100/x. To minimize S, we find S' = 1 - 100/x = 0. Solving gives x = 100, so x = 10. Since x = -10 gives y = -10 and S = -20, while x = 10 gives y = 10 and S = 20, the minimum sum is -20 (achieved with both numbers being -10).
7. Mean Value Theorem
The Mean Value Theorem states that if f is continuous on [a,b] and differentiable on (a,b), then there exists a number c in (a,b) such that:
Example
Verify that the function f(x) = x - 3x + 2x satisfies the Mean Value Theorem on the interval [0,2] and find all values of c that satisfy its conclusion.
Solution: f is continuous everywhere and differentiable everywhere, so it satisfies the hypotheses. We need to find c such that f'(c) = [f(2) - f(0)]/2 = [(8 - 12 + 4) - 0]/2 = 0.
f'(x) = 3x - 6x + 2 = 0. Using the quadratic formula, x = [6 (36 - 24)]/6 = [6 12]/6 = 1 3/3.
Both c = 1 + 3/3 1.58 and c = 1 - 3/3 0.42 lie in the interval [0,2], so they satisfy the Mean Value Theorem.
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