Math 210 Midterm #1 Review

Transcription

1 Math 20 Miterm # Review This ocument is intene to be a rough outline of what you are expecte to have learne an retaine from this course to be prepare for the first miterm. : Functions Definition: A function f is a rule for assigning to each input a unique output. If the input is enote x, then the output is enote f(x). Calculus is the stuy of a particular sort of function, namely real-value functions of a single real variable. This means that the inputs an outputs of a function are real numbers. In this class, whenever we say function, we always mean a real-value function of a single real variable. Definition: The omain of a function f is the set of inputs of f. The range of a function is the set of outputs of the function. Given two functions f(x) an g(x), there are a number of operations that can be performe on them to get new functions. These inclue arithmetic operations an function composition. So, for example, f + g is a function whose value at x is given by (f + g)(x) = f(x) + g(x). Definition: For a function f, if there exists a function g which has the property that g(f(x)) = x for all x in the omain of f, then we call g the inverse of f an enote the inverse function by f. There are some important facts to keep in min about inverse functions. A function f has an inverse precisely when it is one-to-one, or equivalently, when the graph of f passes the horizontal line test. If the inverse function f exists, then its range is the omain of f an its omain is the range of f. To fin a formula for the inverse function, set y = f(x), swap the positions of x an y to get x = f(y), then solve this equation for y. The solution will be y = f (x). Some functions o not pass the horizontal line test an hence, o not have inverses. If an inverse is esire, what is often one is to restrict the omain of the function to one on which it is one-to-one. This is one with the trig functions. For example, sin(x) is not one-to-one. But if we restrict the omain of sin(x) to [ π 2, π 2 ], then this function is one-to-one. We efine the inverse function sin (x) to be the inverse of this restricte function. Similarly, cos (x) is the inverse of the function cos(x) restricte to [0, π] an tan (x) is the inverse of tan(x) restricte to [ π 2, π 2 ]. 2: Limits an Continuity Definition: If the values of a function f(x) can be mae as close as we like to some number L by making x be sufficiently close to (but not equal to) a, then we write lim f(x) = L an say, the limit as x approaches a of f(x) is equal to L. We also have the notion of one sie limits: lim f(x) an lim f(x). + If lim f(x) = L, then both one-sie limits are equal to L. Conversely, if both one-sie limits are equal to L, then lim f(x) = L. For us, limits ten to get complicate when quotients are involve an the limit of the enominator is equal to 0. In this case, you must use sophisticate analysis. If the limit of the numerator is nonzero, then you know the limit oes not exist. If the limit of the numerator is 0, then we can t say anything without further analysis.

2 One goo strategy is to employ a factor an cancel strategy. This strategy always works in computing limits of rational functions where the limits of the top an bottom both go to 0: Example: Calculate the following limit: lim x x 2 x 2 + 2x 3. Solution: Notice that as x approaches, both the numerator an enominator of the rational function go to 0. So, we shoul apply the factor an cancel strategy. Let s factor both the top an the bottom of the rational function, then cancel: lim x x 2 x 2 + 2x 3 (x + )(x ) = lim x (x + 3)(x ) x + = lim x x + 3 = 2 4 = 2 We factore the top an the bottom of the rational function, notice that they both ha a factor of (x ) in common, cancele them (we are allowe to o so since x, then took the limit of the result which we foun to be 2. Often the temptation in evaluating limits is to just stick the number that x is getting closer to into the function an see what comes out. This oes not always work. The above example is a situation where the limit of the function is not equal to function evaluate at the value x is getting closer to since the function is not even efine there. However, in nice situations, we o have this simple way of evaluating limits in our toolbox. This lovely state of affairs we call continuity: Definition: A function f(x) is sai to be continuous at a point x = a if three conitions hol: f(a) exists (i.e., a is in the omain of f). lim f(x) exists. lim f(x) = f(a). If a function is continuous for all points, we say the function itself is continuous. A very nice way to think about continuity is in terms of the graph of the function. A function f(x) is continuous if its graph can be rawn without lifting up the rawing implement. This implies that graphs of continuous functions o not have holes, vertical asymptotes, jumps, an so on. Most of the functions we use in calculus are continuous. Polynomials, rational functions, the trig functions, exponentials an logarithms are all continuous on their omains. Continuous on their omains means that the only way one of these functions fails to be continuous at a point is if the function is not efine there (the first conition of continuity is broken). 3: Derivatives Now we use the concept of limit to efine one of the central notions of calculus: Definition: The erivative of a function f(x) is a new function enote by an is given by the limit: f (x) or f x or x f(x) f (x) = lim h 0 f(x + h) f(x) h The erivative measures two things (which are really equivalent to one another if unerstoo eeply): The erivative of a function f(x) at a particular point x = a, that is, f (a), is the slope of the tangent line to the graph of f(x) at x = a.

3 The erivative of a function f(x) at a particular point x = a, that is, f (a), measures the instantaneous rate of change of the function f(x) when x = a. It is important to unerstan both of these facts. Taking erivatives of functions is really taking limits. However, we can use various results to greatly simplify the process of taking erivatives an reuce it to a fairly mechanical process. Here are the results of this sort that you have learne thus far (you must know these if you hope to have any chance of not being burnt toast on this exam): Constant Rule: x c = 0 Power Rule: Constant Multiple Rule: x xn = nx n x c f(x) = c x f(x) Sum Rule : (f(x) + g(x)) = f (x) + g (x) Prouct Rule : Quotient Rule : Chain Rule : (uv) = u v + v u ( u v ) = u v v u v 2 f x = f u u x Derivatives of Inverse Functions: x f (x) = f (f (x)) Notice that I use many ifferent notations for erivatives in the above rules. It is important to be comfortable with them all since ifferent notations are useful in ifferent contexts. These rules alone are sufficient to compute the erivatives of functions involving powers an arithmetic operations, such as rational functions an functions with raicals. To expan the class of functions we can ifferentiate, you nee to know the erivatives of the trig functions. You shoul just memorize these: sin(x) = cos(x) x cos(x) = sin(x) x Example: Compute the erivative of x tan(x) = sec2 (x) x cot(x) = csc2 (x) f(x) = x2 sin(x) x cos(3x). sec(x) = sec(x) tan(x) x csc(x) = csc(x) cot(x) x Solution: Since f is a quotient, we will use the quotient rule. let u = x 2 sin(x) an v = x cos(3x). The quotient rule emans that we calculate u an v. First, let s compute u. Note that u is itself a prouct of x 2 an sin(x), so we must use the prouct rule to fin u. So we get: u = 2x sin(x) + x 2 cos(x). To compute v, we use the sum rule. Note that x can be rewritten as x 2 an we can take the erivative of this using the power rule. Note also that cos(3x) is a composite function, so we must use the chain rule to fin its erivative. Putting all this together: Then, by the quotient rule: v = 2 x 2 ( 3 sin(3x)) = 2 x + 3 sin(3x)

4 f x = (2x sin(x) + x2 cos(x))( x cos(3x)) ( 2 x + 3 sin(3x))(x2 sin(x)) ( x cos(3x)) 2 If you unerstan this example, you will be well prepare for the questions about ifferentiation on the miterm. However, this problem is not representative of what you might see on the miterm it s too complicate. Example: Suppose y = sin(x) x. (a.) Fin y x. (b.) Note that the point (π, 0) is a point on the curve escribe by the above equation. Base on your work in part (a.), what is the slope of the tangent line? (c.) Base on your work in part (b.), fin the equation of this tangent line. Solution: (a.) Quotient rule: Let u = sin(x) an v = x. Then u = cos(x) while v =. Therefore: y x cos(x) sin(x) = x x 2 (b.) The slope is given by the value of the erivative of y at x = π: y(π) = π cos(π) sin(π) π 2 = π π 2 = π (c.) So we know the slope. Using the fact that the line passes through the point (π, 0) allows us to pinpoint the line: y = π x + b 0 = π π + b Thus, the line is given by: b = y = π x + Suppose you have a curve in the plane efine by an equation which oes not allow y to be expresse explicitly as a function of x. For example, the unit circle is just such a curve: x 2 + y 2 = Since this curve is not the graph of any function, we cannot expect to be able to write y explicitly in terms of x. Nonetheless, we might still be intereste in quantities like slopes of tangent lines to the curve. This is where implicit ifferentiation comes in. This is a technique which will uncover a erviative y = y from an equation which x efines y implicitly. The iea is this: start with the equation, ifferentiate both sies with respect to x being minful to use the all-important chain rule whenever you can, then try to solve for y if possible. Let s o an example: Example: Compute y where 3y 2 + cos(y) = x 3. Solution: We begin by ifferentating both sies with respect to x. Let s take a close look at this process by ifferentiating 3y 2. We use the chain rule here. We first take the erivative of 3y 2 with respect to y, then multiply by the erivative of y with respect to x. Please rea this sentence 0 times if it seeme mysterious to you. What we en up with is 6y y x. In the problem, after ifferentaiation we get: 6y y y sin(y) x x = 3x2

5 Why o we see that y x after the sin(y)? Why on t we see it after the 3x2? I expect you to know why an be able to accurately reprouce the above line on the miterm an final. The next step in the problem is to solve for y. This is just algebra: x y (6y sin(y)) = 3x2 x y x = 3x 2 6y sin(y)