MTH405_Wk_8_Lect_2 - MTH405 Wk 8 Lect 2 Gradient Function f...

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Unformatted text preview: MTH405 Wk 8 Lect 2 Gradient Function f 0 (x) and Tangent Lines Denition. Given a function f (x) on real numbers. Then the derivative, f 0 (x), is the gradient function of f (x), which takes input x values and gives as output the gradient (or slope) of the curve at the coordinate x. Note that the gradient of f (x) at x is exactly the gradient of the tangent line touching f (x) at x. Example. Given the function f (x) = x2 + x, determine 1. the value of f (x) when x = 2 and x = 5. 2. the gradient function of f (x). 3. the values of the slopes of f (x) at x = 2 and x = 5. 4. the equation y = mx + c of the tangent line touching the curve at x = 2. 1 Critical Points Denition. Given a function f (x) on real numbers and c a point on f (x). Then c is a critical point (or stationary point) on f if one of the two following conditions is satised: 1. either f 0 (x) = 0 when x = c (turning point or vertex) or 2. f 0 (c) becomes undened (asymptote). Remark. Two points to note: • A turning point for a function is simply the vertex of the function. • For a turning point x0 , the tangent line at that point is horizontal, hence the gradient of the tangent is 0. This means the gradient function f 0 (x) is also 0. (That's why f 0 (x) = 0 condition). 2 Example. Given a cubic function below. • Note that the gradient of the function f (x) at Point P = (x0 , y0 ) and Q = (x1 , y1 ) are 0. Or in other words, f 0 (x0 ) = 0 f 0 (x1 ) = 0 . • Hence Point P and Q are critical points of f . • For Point P , since it is the maximum point for the immediate region around itself, it is called relative maximum (or local minimum). • For Point Q, since it is the minimum point for the immediate region around itself, it is called relative minimum (or local maximum) . • Both the relative maximum and relative minimum together are called relative extrema . • Note that the cubic function f (x) is dened in the interval from Point O to Point R. • Point O is the absolute minimum since it is the absolute lowest point of f (x) in the interval. • Point R is the absolute of f (x) in the interval. • The absolute minimum and absolute maximum together are called the absolute extrema. maximum 3 since it is the highest point First Derivative Test (FDT) To nd the critical points and to classify them as relative maximum or relative minimum, we use the First Derivative Test Steps: 1. 2. Find all the critical points of f (x) by equating f 0 (x) = 0 and solve for x. Then use the Sign Test using f 0 (x) to determine the nature of the critical point x0 . (a) (b) (c) If the values of f 0 (x) are positive on the left of x0 and negative on the right of x0 , then x0 is a relative maximum. If the values of f 0 (x) are negative on the left of x0 and positive on the right of x0 , then x0 is a relative minimum. If the values of f 0 (x) are positive (or negative, resp.) on the left of x0 and positive (or negative, resp.) on the right of x0 , then x0 is not a relative extrema. 4 Example. For the function f (x) = 4x2 − 12x + 10, nd all critical numbers and determine whether each represents a relative maximum, relative minimum, or neither. Then nd the absolute extrema on the interval [−2, 3]. Solution. We rst determine the derivative of f (x). f 0 (x) = 8x − 12 (gradient function). The equating to 0, we have 0 = 8x − 12 3 x0 = . 2 Hence x = 3 2 is a critical point of f (x). Then using the Sign Test, we take two points: x = 1 on the left and x = 2 on the right of x0 and input in f 0 (x) f 0 (1) = −4 (negative) f (2) = 4 (positive). Since we have negative gradients on the left and positive gradients on the right, by FDT we conclude that x0 = 32 is a relative minimum. To nd absolute extrema, we take the end points of the interval (x = −2, 3) and relative extrema (x = 23 ) and nd their y -values. f (−2) = 50 f (3) = 10   3 f = 1 2 Using the y -values, we determine the absolute extrema. Hence, x = −2 is the absolute maximum and x = 32 is the absolute minimum. 5 ...
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