U3L3: Sec.3.2 Minimum and Maximum on an Interval (Extreme Values)

U3L3: Sec.3.2 Minimum and Maximum on an Interval (Extreme Values)

<p> MCV4U Sec.3.2 Note U3L3: Sec.3.2 Minimum and Maximum on an Interval (Extreme Values)</p><p>Grade 11: Determine the vertex of y  x2  6x 19 . This function is in standard form. We need to complete the square to get it to vertex form. Show using algebra tiles and with algorithm. y  x  32 10</p><p>Therefore the vertex is ( 3 , 10 ). The minimum value of the function would be 10 when x = 3. The min./max. value is the y-portion of the vertex and where this min./max. occurs is the x-portion of the vertex.</p><p>Calculus: Determine the max./min. value of f x  x 2  6x 19 and state where it occurs.</p><p>Since the min./max. value occurs at the vertex of the function which is a turning point, the slope of the tangent (ie: the derivative) will be zero.</p><p> f ‘(x) = 0 at a maximum or minimum.</p><p> f x  x 2  6x 19 f 'x  2x  6 f x  x 2  6x 19 Therefore, 0  2x  6 this is where it occurs, thus f 3  32  63 19 6  2x f 3  10 3  x</p><p>So the max./min. occurs at x = 3 and is 10 but how do we find out whether it’s a maximum or minimum without relying on previous parabola knowledge. Ie: leading coefficient is + or -.</p><p>At a minimum, f ‘(x) < 0 to the left of f ‘(x) = 0 and f ‘(x) > 0 to the right and vice versa for the maximum. Draw this scenario.</p><p>Create a First Derivative Chart Interval f 'x  2x  6 x < 3 - x = 3 0 x > 3 +</p><p>Therefore as you approach x = 3 from below it is negative (downward slope) and from above it is positive (upward slope) so f(x) has a minimum value of 10 at x = 3.</p><p>Page 1 of 3 MCV4U Sec.3.2 Note</p><p>Local vs. Absolute Maximum/Minimum</p><p>Eg.1: Determine the maximum or minimum values of f x  x3  x 2 and determine where they occur.</p><p>At a max./min. f ‘(x) = 0 3x2 – 2x = 0 x(3x - 2) = 0</p><p>Therefore x = 0 and x = 2/3 </p><p>Interval f 'x  3x2  2x x < 0 + x = 0 0 2 - 0 < x < 3 2 0 x = 3 2 + x > 3</p><p>Therefore f(x) has a local maximum at x = 0 and a local minimum at x = 2/3.</p><p>Page 2 of 3 MCV4U Sec.3.2 Note</p><p>Homework: p. # 1 , 2 , 4 , 5a , 7 , 8 , 11 , 12</p><p>Page 3 of 3</p>

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