AP Calculus BC Unit 5 Analytical Applications to Differentiation

Unit 5: Analytical Applications to Differentiation ● Mean Value Theorem (MVT): If a function is continuous over the closed interval [a, b] and differentiable over the open interval (a, b), then there’s a point between ‘a’ and ‘b’ where the instantaneous rate of change equals the average rate of change over the interval. ○ ○ In terms of a graph, the MVT guarantees a point in which the tangent line of a graph at that specific point is equal to the slope between the endpoints of the graph. ● Extreme Value Theorem: If a function is continuous over a closed interval, then it must have a minimum and maximum value over that interval. ○ Minimum value: The smallest value of a function ○ Maximum value: the highest value of a function ● Critical Point: A point in a function where the first derivative at that ‘x’ value is equal to zero, or does not exist ○ To find the critical points of a function when given an equation of the function, find the first derivative of the function, and then determine the ‘x’ values where f’(x) = 0 or doesn’t exist ○ To find the critical points of a function when given a graph of the function, determine the ‘x’ values where the tangent line at that point would be horizontal, indicating a first derivative of 0, or where the first derivative at that point is nonexistent, either because the tangent line is vertical, or because the graph at that point has a cusp, or sharp edge. ● Local/Relative Extrema: You say that a function has a local/relative maximum or a local/relative minimum at a certain ‘x’ value, which means that the graph at that point has a horizontal tangent line. ○ *This doesn’t mean that the function has its highest/lowest value at that point (that’s the absolute maximum/minimum, which will be described later on. A local extremum can be an absolute max/min, but doesn’t have to) ○ To find the local maximums and minimums of a graph: ■ 1) find the critical points of that function ■ 2) determine the concavity at each critical point of that function by taking the second derivative of the function. ● If f’’(x) < 0, then the function has a local maximum at that point ● If f’’(x) > 0, then the function has a local minimum at that point ○ **note that all local extrema are critical points, but not all critical points are local extrema. ■ It’s possible for a critical point to have neither a local minimum or maximum, if f’’(x) doesn’t change concavity at that point. ● How to know if a function is increasing or decreasing: ○ A function is increasing if f’(x) > 0 at a certain point. ○ A function is decreasing if f’(x) < 0 at a certain point. ○ To determine the intervals in which a function is increasing/decreasing, find the critical points of a function, and then choose an ‘x’ value within an interval between 2 critical points or between a critical point and an endpoint, and plug ‘x’ into f’(x). Then determine if f’(x) is negative or positive, and that'll tell you whether the function is increasing or decreasing within that interval. ● Absolute extrema: the highest/smallest value of the function ○ Can only occur at critical points or endpoints ○ To find the absolute maximum or absolute minimum of a function: ■ 1) find the critical points of a function ■ 2) create a chart with one column for your ‘x’ values and one column for the corresponding f(x) value ■ 3) Fill in the ‘x’ values column with the endpoints of the function and all critical points ■ 4) solve for f(x) for each point in the ‘x’ values column, and the absolute maximum is the value of f(x) that’s the greatest out of all the other values, and the absolute minimum is the value of f(x) that’s the smallest out of all the other values. ● Absolute extrema are the ‘x’ points at which the absolute values occur ● Concavity: ○ As mentioned earlier, concavity of a function is determined by the second derivative. ■ If f’’(x) < 0, then the graph is concave down ■ if f’’(x) > 0, then the graph is concave up. ○ Concavity can also be measured by determining if the derivative of a function is increasing/decreasing at a point. ■ If f’(x) is increasing, then it’s concave up ■ If f’(x) is decreasing, then it’s concave down ○ Points of Inflection: a point of inflection is a point in a function where concavity changes at that point. ■ f’’(x) changes signs (either from positive to negative or negative to positive) ■ To find the point of inflection of a function, find the points where the second derivative is equal to zero. Then, determine the concavity of the intervals between the zeroes and the endpoints. If the second derivative changes signs between intervals, then there’s a point of inflection there. ● Connecting the function, its derivative, and second derivative (graphs) ○ As mentioned before, the derivative of a function is basically the slope of that function. Thus, whenever the derivative is positive, that means the slope of the function is positive, and the function is increasing. Likewise, if the derivative is negative, that means the slope of the function is negative and the function is decreasing. ■ When given a graph of f’(x), you can relate this graph to f(x) and f’’(x). On whatever intervals f’(x) is ABOVE the x-axis, that means the f(x) has a POSITIVE slope and is increasing on those intervals. Similarly, if f’(x) is BELOW the x-axis, then you know that f(x) is decreasing and has a NEGATIVE slope on those intervals. The graph of f’(x) can also tell you the key features of f’’(x), in the same way that the graph of f(x) tells you key features of f’(x). ● Basically, if f’(x) has a positive slope in an interval, that means that the graph of f’’(x) should be above the x-axis in that interval. ■ The graphs of any derivative of a function can be related to each other. ● Optimization Problems: problems that ask you to find the “optimal”, or best, value of a quality ○ The derivative of a function can be used to find the optimal value. ○ Steps to solving optimization problems: ■ 1) Determine the equation for the function that needs to be optimized. ■ 2) Express the function with one variable, by using substitution based off of the information given to you by the problem. ■ 3) Take the derivative of the function and set it equal to 0. ■ 4) Solve for the variable to determine the optimal value of the function. ○ Ex. Find the maximum area of a rectangle whose perimeter is 50 𝑚2 . ■ 1) Area = Length x Width → A = LW ■ 2) 2W = 50 - 2L → W = 25 - L → A = L(25-L) = 25L - 𝐿2 ■ 3) dA/dL = 25 - 2L → 0 = 25 - 2L ■ 4) L = 25/2 ■ 5) W = 25 - L → W = 25 - 25/2 = 25/2 ; A = LW → A = (25/2)(25/2) = 625/4 𝑚2 ● *note how the problem asks you to find the maximum AREA. steps 1-4 only solved for the value of the LENGTH of the rectangle. The rest of the steps vary depending on what the problem’s asking.