How to Sketch a Graph in Calculus
- 1). Find the domain of the function by determining where the function does not exist. This can be at an asymptote or other discontinuity on the function's graph. For example, the function f(x) = cosx / (2 + sinx) has a domain including all real numbers because there are no values where the denominator equals 0.
- 2). Determine the intercepts of the graph by solving the function for x = 0 and f(x) = 0. For example, f(0) = cos(0) / (2 + sin(0)) = (1 / 2), therefore the y-intercept equals 1/2. cosx / (2 + sinx) = 0 when cosx = 0. This occurs when x = (2n + 1)PI / 2, where n is any integer. Therefore, there are an infinite number of x-intercepts.
- 3). Determine the symmetry of the function. If f(x) = f(-x) then the function is even and symmetric about the y-axis. If f(-x) = -f(x) then the function is odd and symmetric about the origin. The function f(x) = cosx / (2 + sinx) is neither even nor odd, so it is not symmetric.
- 4). Find any asymptotes of the function. A horizontal asymptote occurs if the (limit of f(x) as x ---> infinity) approaches a number L, then L is a horizontal asymptote. If the (limit of f(x) as x ---> C) approaches infinity, where C is any number, then C is a vertical asymptote. For example, (x) = cosx / (2 + sinx) has no asymptotes based on these rules.
- 5). Determine the intervals of increase and decrease by using the increase/decrease test for derivatives. This test states that where f '(x) is positive f(x) is increasing and where f '(x) is negative f(x) is decreasing. For example, the derivative of (x) = cosx / (2 + sinx) = -(2sinx + 2) / (2 + sinx)^2, by the Quotient Rule of derivatives. Thus, f '(x) > 0 when sinx < -1/2 or when (7PI / 6) < x < (11PI / 6) so f(x) is increasing here. So, it decreases on the intervals (0, 7PI / 6) and (11PI / 6, 2PI). 0 to 2PI is the period of the sine and cosine functions.
- 6). Find the critical numbers of f(x) in order to determine any maximum or minimum values. Critical numbers of defined as any number c where f '(c) = 0. If f '(c) changes sign from positive to negative at c then f(c) is a maximum and if f '(c) changes sign from negative to positive at c then f(c) is a minimum. For example, substituting the end points of the intervals of the increase/decrease test, it is found that a minimum occurs at f(7PI / 6) = -1 / sqrt(3) and a maximum occurs at f(11PI / 6) = 1 / sqrt(3).
- 7). Find the second derivative of the function, f ''(x), to determine points of inflection and concavity. The second derivative is the derivative of f '(x). For example, using the rules of differentiation, the derivative of f '(x) = -(2sinx + 2) / (2 + sinx)^2 is f ''(x) = -(2cosx(1 - sinx) / (2 + sinx)^3. When f ''(x) is positive then f(x) is concave up. When f ''(x) is negative then f(x) is concave down. Examining f ''(x) finds that it is positive when cosx < 0 ---> (PI / 2) < x < (3PI / 2). So f(x) is concave up on the interval (PI / 2, 3PI / 2) and concave down on (0, PI / 2) and (3PI / 2, 2PI). The points of inflection are defined as any points on the graph of f ''(x) where there is a change of signs.
- 8). Sketch the graph using all information discovered from using calculus on the function. First plot x- and y-interecepts, maximum and minimum points and inflection points. Asymptotes should be drawn as dotted lines. As you draw the graph, pass it through the plotted points based on the intervals of increase and decrease. Pay attention to the direction of concavity determine by the behavior of f "(x).
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