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Basic Calculus
LIMITS OF FUNCTIONS USING TABLE OF VALUES AND THE
GRAPH
CALCULATING LIMITS
Limits are tool for reasoning about function behavior, and tables are a tool reasoning about limits. 𝑙𝑖𝑚 𝑓 ( 𝑥 ) = L 𝑥 → 𝑐 LIMIT OF A FUNCTION USING TABLE OF VALUES The limit of a function is real number 𝐿 that 𝑓(𝑥) approaches a given number c, written 𝑙𝑖𝑚 𝑓(𝑥) = L 𝑥→𝑐 Read as “The limit of 𝑓(𝑥) as 𝑥 approaches 𝑐 is 𝐿”. The limit of the function from left and right are equal. Therefore, 𝑙𝑖𝑚 (2𝑥 + 1) = 5 𝑥→ ➢One sided limit is the value L as the x value gets closer and closer to a certain value of c from one side. 𝑙𝑖𝑚 𝑓(𝑥) = 𝐿 From the right 𝑥→𝑐+ 𝑙𝑖𝑚 𝑓(𝑥) = 𝐿 From the left 𝑥→𝑐− ➢If the limit of the function from the left is NOT equal to the limit from the right, then the limit DOES NOT EXIST. ➢The limit of a function 𝑙𝑖𝑚 𝑥→𝑐 𝑓(𝑥) = 𝐿 and 𝑓(𝑐) are not always the same. LIMIT OF A FUNCTION USING GRAPH EXAMPLE 1: Consider the graph of 𝒇 𝒙 = 𝟐𝒙 + 𝟏 as shown at the right. The limit of the function from the left is 2 and from the right is also equal to 2. The limit of the function is 2. EXAMPLE 2 : 𝑙𝑖𝑚 𝑓(𝑥) = 1 𝑥→1+ 𝑙𝑖𝑚 𝑓(𝑥) = − 𝑥→1− 𝑙𝑖𝑚 f(𝑥) does not exist (DNE) because 𝑥→1 limit from the left side and limit from the right side are not equal.
LIMIT LAWS
- The limit of a constant is itself. If 𝒌 is any constant, then 𝒍𝒊𝒎 𝒌 = k 𝒙→𝒄 Examples: 𝑎. 𝑙𝑖𝑚 5 = 5 𝑥→𝑐
- The limit of x as x approaches c is equal to c. 𝑙𝑖𝑚 x = c 𝑥→c Example: (a) 𝑙𝑖𝑚 x = - 1 𝑥→− (b) 𝑙𝑖𝑚 x= 10 𝑥→ (c) 𝑙𝑖𝑚 x = 0 𝑥→ For the remaining theorems, we will assume that the limits of f and g both exist as x approaches c and that they are L and M respectively, 𝑙𝑖𝑚 f(x) = L and 𝑙𝑖𝑚 g(x) = M 𝑥→c 𝑥→c
- The Constant Multiple Theorem : the limit of a constant times a function is equal to the product of the constant and the limit of the function.
- The Addition/Subtraction Theorem : the limit of a sum/difference of functions is the sum/difference of the limits of the individual functions.
- The Multiplication Theorem: the limit of a product of functions is equal to the product of their limits.
- The Division Theorem: the limit of a quotient of functions is equal to the quotient of the limits of the individual functions, provided the denominator limit is not equal to zero.
Enlarge illustration of the coordinates plotted on a Cartesian Plane EXPONENTIAL FUNCTION
TRIGONOMETRIC FUNCTION
SINE
Enlarge illustration of the coordinates plotted on a Cartesian Plane.
COSINE
NOTE:
CONTINUITY AND DISCONTINUITY
CONTINUITY OF A FUNCTION
There are different ways on how we can represent a function. One of those is its graph. If the graph of a function has NO GAP or HOLES, then we can say that the function is CONTINUOUS. Otherwise, it’s DISCONTINUOUS. In addition, a function is said to be continuous at a point 𝒙 = 𝒂 if all of the following conditions are satisfied (Comandante, 2008):
- If 𝒇(𝒙) is defined, that is, exists at 𝒙 = 𝒂.
- A number exists if it is defined. Division by zero yields to undefined values. The same with square root of negative numbers. Thus, these cases must be checked.
- If 𝒍𝒊𝒎 𝒇(𝒙) exists. 𝒙→𝒄
- The limit of a function 𝒇(𝒙) exists if the left-hand limit is equal to the right-hand limit. That is, 𝑙𝑖𝑚 𝒇( 𝒙) = 𝑙𝑖𝑚 𝒇( 𝒙) 𝒙→𝒄− 𝒙→𝒄+
- If lim 𝑓(𝑥) = 𝑓(𝑐) 𝑥→𝑎
- Upon checking whether a function satisfies the first two conditions, check whether the obtained values from these conditions are equal to each other.
Since the two values are equal, then the third condition is satisfied. Since all of the three conditions were satisfied, then we can say that the function 𝑓 ( 𝑥 ) = 𝑥 2 + 5 𝑥 + 6 is continuous at 𝑥 = −1. ILLUSTRATION:
EXAMPLE 2:
EXAMPLE 3:
Determine if the function𝑓(𝑥) = 2/ 𝑥 continuous at 𝑥 = 0
CONTINUITY OF A FUNCTION AT A CLOSED INTERVAL [A,B]
A function is said to be continuous at a closed interval [a, b] if its right endpoint, open interval and left endpoint has no breakage, holes or discontinuity. (see figure below)
- The following are the conditions needed to be satisfied to be able to know whether the function is continuous or not on a closed interval.
- The function 𝑓 ( 𝑥 ) needs to be continuous at the right endpoint interval 1.
SLOPE OF THE TANGENT LINE TO A CURVE
TANGENT LINE
- A line is tangent to a circle if it intersects the circle at exactly one point. PROPERTIES OF TANGENT LINES EXAMPLE 1: Tangent lines drawn at the maximum or minimum points are horizontal. EXAMPLE 2: Tangent lines drawn to a line, is the line itself. EXAMPLE 3: Vertical tangents may exist even for continuous functions CASES WHEREIN CURVES DO NOT HAVE TANGENT LINES AT SPECIFIC POINTS EXAMPLE 4; When the function is not continuous at P.
EXAMPLE 5:
When the function is not continuous at P. To find the tangent line at Point P, there is a need for a second point Q on the curve. o If a Point Q will slide down to point P, it will get closer to point P and the slope of secant PQ will then approach the value of the slope of line l tangent to the curve at point P. o This is where the slope of a tangent line is derived. As the difference in the distance in x gets smaller, the slope of the secant line gets closer and closer to the slope of the tangent line. DEFINITION: Let C be the graph of a continuous function 𝑦 = 𝑓(𝑥) and let 𝑃 be a point on 𝐶.
- A secant line to 𝑦 = 𝑓(𝑥) through 𝑃 is any line containing 𝑃 and another point 𝑄 on 𝐶.
- The tangent line to 𝑦 = 𝑓(𝑥) at 𝑃 is the limiting position of all secant lines 𝑃𝑄 as Q → 𝑃. EQUATION OF THE TANGENT LINE
EXAMPLE 1:
EXAMPLE 2 :
Using the same given in Example 1, write the equation of the tangent line at the given point. To write the equation of the line, we may use the point-slope form of the line, 𝑦 − 𝑦 1 = 𝑚(𝑥 − 𝑥 1 )
DIFFERENTIATING AN EXPONENTIAL FUNCTION
DIFFERENTIATING A LOGARITHMIC FUNCTION
DIFFERENTIATION OF TRIGONOMETRIC FUNCTIONS
CHAIN RULE
Chain Rule is the process of differentiating a composite function. Recall: Composite functions are two functions combined to make a single one. For example, the combination of functions 𝑓 and 𝑔 : ( 𝑓 𝑜 𝑔 )( 𝑥 ) = 𝑓 ( 𝑔 ( 𝑥 )) Note: To apply the Chain Rule on composite functions, you must take the derivative of its outside function and then multiply it to the derivative of its inside function.
EXTREME VALUE THEOREM (EVT)
✓ This theorem states that a function 𝑓(𝑥) which is found to be continuous over a closed interval [𝑎, 𝑏] is guaranteed to have extreme values in that interval. ✓ An extreme value of 𝑓 or extremum, is either a minimum or maximum value of a function. ❖ A minimum value of 𝑓 occurs at some 𝑥 = 𝑐, if 𝑓(𝑐) ≤ 𝑓(𝑥) for all 𝑥 ≠ 𝑐 in that interval. ❖ A maximum value of 𝑓 occurs at some𝑥 = 𝑐, if 𝑓(𝑐) ≥ 𝑓(𝑥) for all 𝑥 ≠ 𝑐 in that interval. EXAMPLE 1: Sketch the graph and find the minimum and maximum values of the function 𝑓 𝑥 = 5𝑥 2 + 2𝑥 − 3 at the interval [−3, 2]. Since 𝑓 𝑥 = 5 𝑥 2 + 2 𝑥 − 3 is a quadratic function, its graph is a parabola which opens upward, so its minimum point is its vertex at (0, −3) and the maximum point in the interval [− 3 , 2 ] is ( - 3 , 36 ). Therefore, the minimum value is - 3 and the maximum value is 36.
EXAMPLE 2:
Does the function 𝑓 𝑥 = 𝑥− 1 𝑥+1 at [−4, 4] have extrema? Explain your answer. Notice that the graph of the function breaks since it will be undefined at 𝑥 = −1. Therefore 𝑓 𝑥 = 𝑥− 1 𝑥+1 has no maximum or minimum value because it is not a continuous function.
OPTIMIZATION PROBLEMS
✓ Optimization problems are word problems that deal with the application of finding the maximum or minimum value of a function. It has a constraint and an equation that needs to be optimized. ✓ Constraint is a conditional concept that can be transformed into an equation which is part of an optimization problem. Most problems have a given constant quantity. ✓ Optimization equation is part of the problem that needs to be maximized or minimized. GUIDELINES IN SOLVING OPTIMIZATION PROBLEMS:
- Read and understand the problem carefully.
- Create a sketch or illustration with given labels whenever necessary.
- List what is asked and what are the given data in the problem.
- Write down the constraint and optimization equations.
- Equate one variable in terms of the other from the constraint equation. Then, replace or substitute it to the same variable on the optimization equation.
- Simplify the optimization equation then solve for its first derivative. 7. Set the first derivative equation to zero and find the critical point. 8. Test the critical point values using the derivative tests. 9. Substitute the critical value to solve for the other. EXAMPLE 1: Find two positive numbers such that the sum of the first and twice the second is 100 and their product is a maximum. SOLUTION: EXAMPLE 2: Solve for the dimensions of a rectangle whose perimeter is 250 m with a maximum area. Let 𝑥 and 𝑦 be the missing sides of the rectangle with perimeter 𝑃 = 250 𝑚. SOLUTION:
- Draw a diagram. List what is asked on the problem and label the diagram with relevant data.
