You should prepare the following topics for our final exam. () Pre-calculus. (2) Inverses. (3) Algebra of Limits. (4) Derivative Formulas and Rules. (5) Graphing Techniques. (6) Optimization (Maxima and Minima). (7) Integration Formulas and Rules. (8) The Fundamental Theorem of Calculus. (9) Logarithmic and Implicit Differentiation. () L Hospital s Rule. () Pre-calculus. Know the rules for exponential functions, logarithms, trigonometry, graphing basics and inverses. You should know basic trigonometry, including the graphs, identities and definitions of the trig functions. The exponential laws and the laws of logarithms are very important. These are a x a y = a x+y a x a = y ax y (a x ) r = a rx and log a (xy) = log a (x) + log a (y) log a ( x y ) = log a(x) log a (y) log a (x r ) = r log a (x). The graphing basics that you should know include the following. The domain of is all real numbers where the function is defined. You find the domain by eliminating points which create division by zero or an even root of a negative number. The range of is the set of all values the funtion takes on. You can get the range by describing the output of the function as the input varies over all x in the domain of. The x intercept(s) are the points (x, ),, (x n, ) where x,, x n are found by solving = for x. The y intercept is the points (, f()) where you get the y coordinate by plugging into your function. The graph of cf(x + a) + b is a scale by c of the graph of shifted left a units and up b units. Know the basic graphs like x n for integers n, e x, ln(x) and the trigonometry functions. A function f has the vertical line x = c as an asymptote exactly when f(c) has denominator zero and numerator which is not zero. If lim x c = the function drops down the line x = c on the left of this line. If lim x c = the function climbs up the line x = c on the left of this line. If lim x c + = the function drops down the line x = c on the right of this line. If lim x c + = the function climbs up the line x = c on the right of this line. If = anxn + +a x+a b k x k + +b xb has the form of one polynomial divided by another there a 3 possibilities. If n > k there may be a slant asymptote but there is no horizontal asymptote; if n = k the line y = an b k is a horizontal asymptote; if n < k the line y = (i.e., the x-axis) is a horizontal asymptote. A function can cross a horizontal asymptote. A horizontal asymptote describes the behavior of the function at the left edge (lim x ) and the right edge (lim x ) of the graph.
(2) Inverses: A function is one-to-one if whenever a b, f(a) f(b). A function is one-toone exactly when no horizontal line crosses the graph of in more than one place. Let f : X Y be one-to-one. Then f (x) : Y X is defined by f () = x for all x in X or equivalently f(f (y)) = y for all y in Y. The graph of f (x) is the reflection of the graph of about the line y = x and the derivative of f is given by (f ) () =. f (x) The most important inverse pairs are logarithms and exponentials. We also studied the inverse trig functions sin (x), cos (x), and tan (x). (3) Algebra of Limits. If lim x a If lim x a positive number = then lim x a negative number = then lim x a If lim x a = non-zero number then lim x a If lim x a If lim x a non-zero number = then lim non-zero number x a =. is this number. = then you have to do some algebra to find lim x a. You try to rewrite to remove the division by zero reduce the problem to one of the above cases. (See also: L Hospital s rule.) (4) Derivative Formulas and Rules. Know all derivatives formulas and derivative rules. It is helpful to recall that by definition, f f(a+h) f(a) (a) = lim h. Also remember that the derivative at a is the h slope of the tangent line to the graph of at the point (a, f(a)). The equation of this tangent line is y f(a) = f (a)(x a). Derivative Rules: (Where c is a constant.) (c) = c ( + ) = + g (x) () = + g (x) ( ) = g (x) ( ) = g (x) (f()) = f ()g (x) () 2 Derivatives: (Where c is a constant.) (sin(x)) = cos(x) (cos(x)) = sin(x) (sec(x)) = tan(x) sec(x) (x n ) = nx n (c) = (csc(x)) = cot(x) csc(x) (cot(x)) = csc 2 (x) (tan(x)) = sec 2 (x) (a x ) = ln(a)a x (log a (x)) = (e h(x) ) = h (x)e h(x) (ln(h(x)) = h (x) ln(a)x h(x) (5) Graphing Techniques. We use the first and second derivatives to calculate when a graph is Increasing/Decreasing, Concave up/down, or has a Maxima/Minima/Inflection Point. Recall that f > is equivalent to f is increasing, meaning when a < b, f(a) < f(b). Similarly f < is equivalent to f is decreasing, meaning when a < b, f(b) < f(a). A function has a maximum where it switches from increasing to decreasing and a minimum where it switches from decreasing to increasing. A function f is concave up exactly when f >. This is equivalent to saying f is increasing and so the graph of f will resemble part of a porabola opening upward.
(6) Optimization (Maxima and Minima). The extrema of on [a, b] are the local and global maxima and minima of on [a, b]. To find the extrema: (i) Calculate f (x). (ii) Solve for the x-values where the derivative is zero or does not exist - i.e., the critical values of on [a, b]. (iii) List these critical values in increasing order a = c c 2 c n = b. (iv) Plug the critical values into the function to get f(c ), f(c 2 ),, f(c n ). (v) Classify each c i by comparing f(c i ) to f(c i ) and f(c i+ ). (7) Integration Formulas and Rules. By construction, if > then b dx a is the area below the graph of and above the x-axis between the vertical lines x = a and x = b. The integral is given by reversing the derivative process using the following laws. Anti-derivative Rules: (Where c isa constant.) c dx = c dx + dx = dx + dx () dx = + c dx = dx dx b a a () dx = f(b) f(a) dx = b dx b a Anti-derivatives: (Where c is a constant.) f ()g (x) dx = f() + c b dx = c dx + b dx a a c sin(x) dx = cos(x) + c csc 2 (x) dx = cot(x) + c sec 2 (x) dx = tan(x) + c cos(x) dx = sin(x) + c x n dx = xn+ + c (n ) n+ cot(x) csc(x) dx = csc(x) tan(x) sec(x) dx = sec(x) a x dx = ax + c ln(a) dx = log ln(a)x a(x) + c h (x)e h(x) dx = e h(x) + c h (x) dx = ln(h(x)) + c h(x) Recall that if for all x such that a x b then A = b a dx is the area between the curves and and between the vertical lines x = a and x = b. Recall integration is the method for reversing differentiation. The substitution method of integration is the method for reversing the chain rule. It says, f ()g (x) dx = f (u) du where u = This means to integrate a composition, start with a change of variables. Let u = be the inside function. Then du = dx g (x) and so du = g (x)dx. Substitution into f ()g (x) dx give us f (u) du = f(u) + c then substitute the variable u out and you get the answer f() + c.
(8) The Fundamental Theorem of Calculus The FTC says that the function F (x) = f(t) dt is the anti-derivative of the integrand. Thus, x c F (x) = d dx [ x c f(t) dt] =. Combined with the chain rule, the FTC says that if = h(x) f(t) dt then k(x) g (x) = d dx [F (h(x)) F (k(x))] = f(h(x))h (x) f(k(x))k (x). (9) Logarithmic and Implicit Differentiation. Logarithmic differentiation is used when you need to convert exponents to products or to convert products and quotients to sums and differences. You calculate the derivative f (x) with logarithmic differentiation by (i) Set y =. (ii) Then ln(y) = ln(). (iii) Use the laws of logarithms to expand ln(). (iv) Differentiate both sides using the formula d [ln(] = g (x). This gives you dx y = [Some Mess]. y (v) Multiply both sides of (iv) by y in order to isolate y. Then substitute for y and f (x) for y. We use implicit differentiation when the variables x and y are related by an equation like x 3 + y 3 = 6xy where it is not easy to express y as an explicit function of x. With this method we can find the rate of change in y with respect to x, y = dy, dx without solving for y explicitly. The idea is to recognize that y = y(x), i.e., that y is implicitly defined by x. We take the derivative of both sides of this equation using the chain rule: d dx [f(y)] = d dy [f(y)]dy dx = d dy [f(y)]y. That is, the derivative of a function of y is the usual derivative with respect to y times the chain rule factor y. Using this we can solve for y without ever solving for y in terms of x. () L Hospital s Rule. L hospital s rule is a rule for calculating limits. It says if f(a) g(a) has the form or ± then: lim x a = lim x a f (x) g (x). We can also use algebra to rewrite other special forms until L Hospitals rule does apply. If lim x a h(x) has the form,,, or then h(x) can be algebraically manipulated until it is a limit problem which has the form or ±.
Sample Problems and Solutions () Solve for x in 7 x2 x = 49. ANSWER: There are two approaches: 7 x2 x = 7 2 and so x 2 x = 2. That is, x 2 x 2 = and so (x 2)(x + ) =. This gives the solutions x = 2 and x =. Alternatively, since 7 x2 x = 49, ln(7 x2 x ) = ln(49). That is, (x 2 x) ln(7) = ln(49) which implies that x 2 x = ln(49) ln(7) quadratic equation as above.. Since ln(49) ln(7) = 2 the problem reduces to the same (2) Let = 3x. Find its inverse f (x) and find the Domain, Range and x and y 6x+2 intercepts for both and f (x) ANSWER: Let y = 3x. Then (6x + 2)y = 3x and so 6xy + 2y = 3x. 6x+2 This implies 6xy 3x = 2y or x(6y 3) = 2y. Dividing we get, y = 2y+. 3 6y So, f (x) = 2x+. 3 6x The domain of is all real numbers except and the range of is all real 3 numbers except. The x-intercept of is (, ) and the y-intercept of is 2 3 (, ). 2 The domain of f (x) is all real numbers except. and the range of f (x) is all 2 real numbers except. The x-intercept of f (x) is (, ) and the y-intercept of 3 2 f (x) is (, ). 3 (3) Let = 3x. Sketch the graph of. 6x+2 ANSWER: We know from the previous problem that Domain() is all real numbers except x = and the range of is all real numbers except y =. The 3 2 x-intercept is (, ) and the y-intercept is (, ). 3 2 The function is neither even nor odd. So the graph is not symmetric. It has the vertical asymptote x = and the horizontal asymptote y = 3 =. We calculate 3 6 2 3x lim x ( 6x + 2 = 2 3x = and lim 3 ) x ( 6x + 2 = 2 =. + 3 )+ Thus, climbs up the vertical asymptote its the left side and falls down the vertical asymptote on its right side. We also calculate 3x lim x 6x + 2 = ( 3x 2 )+ and lim x 6x + 2 = ( 2 ). Thus, is above the horizontal asymptote y = on the left edge of the graph and 2 below it on the right edge. Combining all of this, we get that the graph of looks like a shift of y =. It x has a porabola like curve in the top left and bottom right quadrants created by the horizontal and vertical asymptotes.
Sample Problems and Solutions (4) Let = x 4 4x 3 = x 3 (x 4). Sketch the graph of and identify where is increasing/decreasing and concave up/down. Lable all maxima and minima for the function. ANSWER: First note that is a 4th degree polynomial. Its domain is all real numbers. It crosses the x axes at (, ) and (4, ) and the y intercept is the origin. Taking the derivative of = x 4 4x 3 twice we get that f (x) = 4x 3 2x 2 = 4x 2 (x 3) and f (x) = 2x 2 24x = 2x(x 2). The critical values of are and 3. The derivative is negative if x < 3 and positive if x > 3. Thus is decreasing on (, 3) and increasing on (3, ). The second derivative is zero at x = and x = 2 and it alternates signs as positive, negative and the positive on (, ), (, 2) and (2, ), respectively. Thus, has inflection points at x = and x = 2. It is concave up on (, ) and (2, ) and concave down on the interval (, 2). To graph f, we calculate the points on the graph (, 5), (, ), (, 3), (2, 6), (3, 27), and (4, ). The graph looks like a W. It has an absolute minimum at (3, 27). ln(x) (5) Find lim x x ANSWER: The limit has the form and so L hospital s rule implies that lim x ln(x) = x x lim x =. (6) Find lim x (ln(x)) x. ANSWER: This has the form and so we can use the natural log to make this a L hospital s rule problem. Let y = (ln(x)) x. Then ln(y) = ln(x). Now lim x x ln(x) x has the form and so by L hospital s rule ln(x) x lim x = lim x x =. This means the original limit is e =. t (7) Find lim 3 9t 2 +27t 27 t 3 t 3 (t 3) ANSWER: You can factor and cancel as before. That is lim 3 t 3 t 3 3) 2 =. You can also use L hospital s rule which gives us lim t 3 t 3 9t 2 +27t 27 t 3 = lim t 3 3t 2 8t+27 =. = lim t 3 (t cot(x) (8) Find lim x π x ANSWER: This has the form exist as a finite number). cos(π) = sin(π)(π). So the limit is infinite (i.e., does not (9) Find the equation of the tangent line to the graph at = x 7 + 4x at x =. ANSWER: The line goes through (x, f(x )) = (, 5). The slope is f () = 7() 6 + 4 =. The tangent line is y 5 = (x ) or y = x 6.
Sample Problems and Solutions () Find the derivative of = (3x+5)7 (4x 3) 9. (6x+2) 2 ( 3x+8) ANSWER: You can use the quotient rule to get the derivative but the algebra gets complicated. A better option is logarithmic differentiation. Let y = (3x+5)7 (4x 3) 9. (6x+2) 2 ( 3x+8) Then ln(y) = 7 ln(3x + 5) + 9 ln(4x 3) 2 ln(6x + 2) ln( 3x + 8). Taking the y derivative of both sides of this equation we get: = 2 + 36 2 + 3. y 3x+5 4x 3 6x+2 3x+8 Multiplying both sides of this equation by y gives us: y = ( (3x+5)7 (4x 3) 9 )( 2 + 36 2 + 3 ). (6x+2) 2 ( 3x+8) 3x+5 4x 3 6x+2 3x+8 () Find the derivative of = x 3 +x 4x cot(t) dt. sin(x) ANSWER: g (x) = cot(x 3 + x)(3x 2 + ) cot( 4x 4 sin(x) 4x cos(x) )( ). sin(x) sin 2 (x) (2) Calculate the function F (x) = x π sec 2 (t) dt and then calculate its derivative F (x). 4 Confirm your answer using the Fundamental Theorem of Calculus. ANSWER: Integrating we get F (x) = x sec 2 (t) dt = tan(t) t=x t= π 4 = tan(x) tan( π ) = tan(x). 4 Taking the derivative of this we get F (x) = sec 2 (x). Alternatively, the FTC gives us that F (x) = d x dx [ sec 2 (t) dt] = sec 2 (x)() sec 2 ( π 4 )() = sec2 (x). The factors () and () are the derivatives of the upper and lower limits of integration, respectively. (3) Find the total area between the graph of = x 3 and the x-axis between the vertical lines x = 2 and x = 3. ANSWER: The graph of is just the graph of x 3 shifted down unit. From this picture we see that > when < x < 3 and < when 2 < x <. This means the total area is 2 x3 dx + 3 x3 dx. We calculate these integrals separately. 2 3 x 3 dx = x4 4 x 2 = 4 4 2 = 63 4 x 3 dx = x4 4 x 3 = 8 4 3 4 + = 8. So the total area is 6 3 + 8 = 24 3. 4 4 Instead of graphing you can also find the place x = where the functions changes signs by solving =. In this case x 3 = implies x 3 = and so x =.