How to write a good justification in AP Calculus AB The Intermediate Value Theorem (IVT) The IVT is used to prove the existence of some specified y value on a given domain. If a function is continuous on a closed interval , then we can use the IVT to show that the function must pass through all of the y values between the two endpoints. Example: Let f be the function given by f (x) = −e x cos x . Show (without using a calculator) that there exists at least one zero on the interval [0, π ] This function is continuous on the closed interval [0, π ] so IVT must apply. f(0) = -1 so, f(0) < 0 and f( π ) = e π so, f( π ) > 0 Here is the justification: By the IVT, there is a c where, 0 < c < π , such that f(0) < f(c) < f( π ). Since f(0) = -1 and f( π ) = e π , then -1 < f(c) < e π so there must exist an f(c) = 0 for 0<c< π. You should follow this format! The Mean Value Theorem (MVT) Your justification statement should look like: “Since the function is continuous on the closed interval [a, b], and differentiable on the open interval (a, b), then by the Mean Value Theorem there is a c, where a < c < b, such that f (b) − f (a) = f '(c) ”. b−a *Be specific. Do not just state the general rule. Make sure to provide all the specific values for a and b in the justification. Rolle’s Theorem (A ‘special’ case of MVT) You can always use the MVT however if you want to use Rolle’s then the justification statement should look like: “By Rolle’s theorem, since f is continuous on [a, b] and differentiable on (a, b) AND f(a) = f(b), then there is a c where, a < c < b, such that f '(c) = 0 ”. *Once again, be specific. Use the values given in the problem in your justification. Continuity To prove that a function is continuous at x = c, you must show 3 things: 1. f(c) must be defined. 2. lim f (x) must exist and 3. f (c) = lim f (x) x→c x→c #% 2 Example: Let f(x) be the piecewise function described as f (x) = $ x + x +1, x ≤ 2 %& 11− 2x, x > 2 show that f(x) is continuous at x = 2. Solution: f(2) = 7. Also the left hand limit lim f (x) = 7 and the right hand limit x→2− lim f (x) = 7 so lim f (x) = 7 . Since f(2) = 7 = lim f (x) , then f(x) is continuous at x = 2. x→2+ x→2 x→2 Relative/Local Maxima and Minima (extrema) Let x = a be a critical value for f(x). In other words, f '(a) = 0 OR f '(a) is undefined but f(a) is defined. To show that (a, f(a)) is a relative/local minimum: State: “At x = a, f '(a) changes from negative to positive values. Hence, f has a relative minimum at x = a “. OR “ At x = a, f "(a) is a positive value, hence the graph is concave up and the critical value at x = a is a minimum.” To show that (a, f(a)) is a relative/local maximum: State: “At x = a, f '(a) changes from positive to negative values. Hence, f has a relative maximum at x = a “. OR “ At x = a, f "(a) is a negative value, hence the graph is concave down at x = a and the critical value at x = a is a maximum.” A sign chart or graph is not sufficient for justification! A statement must be made. Points of Inflection Let f "(a) = 0 or f "(a) be undefined but f(a) is defined. To show that (a, f(b)) is a point of inflection you must justify with on of the following statements (with the constant replaced with numbers of course): “At x = a, f "(a) changes from positive to negative values (or negative to positive values). Hence, f has a point of inflection at x = a.” If a graph of the first derivative is given, then a point of inflection occurs at x = a if at x = a the graph of f '(x) changes from either increasing to decreasing OR decreasing to increasing. Absolute Maximum or Absolute Minimum The candidates are the critical values of f (x) AND the endpoints. First, justify any relative max/min AND find their function values. Then compare these function values to the endpoint values and decide which is the absolute max/min. Horizontal Tangents occur at x = a if f '(a) = 0 . Vertical Tangents occur at x = a if f '(a) is undefined but f(a) is defined. If f '(x) is of the form f '(x) = g(x) , then f will have a horizontal tangent when h(x) g(x) = 0 AND f will have a vertical tangent if h(x) = 0, provided f(x) is defined at those values. Horizontal Asymptotes y = b is a horizontal asymptote of the graph of y = f (x) if either lim f (x) = b OR x→∞ lim f (x) = b . x→−∞ Limits of Rational Functions as x → ±∞ [end behavior] f (x) = 0 if the degree of f(x) < the degree of g(x) x→±∞ g(x) f (x) lim = ∞ or “dne” if the degree of f(x) > the degree of g(x) x→±∞ g(x) f (x) is finite if the degree of f(x) = the degree of g(x). In this case there will lim x→±∞ g(x) lim be a horizontal asymptote. Increasing/Decreasing and Concavity (Assuming the function is continuous and differentiable) If f '(x) > 0 for every x in (a, b), then f (x) is increasing on [a, b]. If f '(x) < 0 for every x in (a, b), then f (x) is decreasing on [a, b]. If f ''(x) > 0 for every x in (a, b), then f (x) is concave up on (a, b) AND f '(x) is increasing on [a, b]. If f ''(x) < 0 for every x in (a, b), then f (x) is concave down on (a, b) AND f '(x) is decreasing on [a, b]. If f '(x) is increasing on (a,b), then f(x) is concave up on (a, b) AND f ''(x) > 0 on (a, b). If f '(x) is decreasing on (a,b), then f(x) is concave down on (a, b) AND f ''(x) < 0 on (a, b). Motion If s(t) is the position function, then Velocity, v(t) = s’(t) and Acceleration, a(t) = v’(t) = s’’(t). Speed is v(t) . To justify that the speed of a particle is increasing at x = c you need to clearly state that v(c) > 0 AND a(c) > 0 OR that v(c) < 0 AND a(c) < 0. To justify that the speed of a particle is decreasing at x = c you need to clearly state that v(c) > 0 AND a(c) < 0 OR that v(c) < 0 AND a(c) > 0. Optimization You should always use calculus. Find a primary equation that fits what you are trying to optimize. If necessary, find a secondary equation that will allow you to rewrite your primary equation in terms of one variable. Then use calculus techniques to find the maximum or minimum.