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

MCV4U Sec.3.2 Note U3L3: Sec.3.2 Minimum and Maximum on an Interval (Extreme Values)

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

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.

Calculus: Determine the max./min. value of f x  x 2  6x 19 and state where it occurs.

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.

f ‘(x) = 0 at a maximum or minimum.

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

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 -.

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.

Create a First Derivative Chart Interval f 'x  2x  6 x < 3 - x = 3 0 x > 3 +

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.

of 3 MCV4U Sec.3.2 Note

Local vs. Absolute Maximum/Minimum

Eg.1: Determine the maximum or minimum values of f x  x3  x 2 and determine where they occur.

At a max./min. f ‘(x) = 0 3x2 – 2x = 0 x(3x - 2) = 0

Therefore x = 0 and x = 2/3

Interval f 'x  3x2  2x x < 0 + x = 0 0 2 - 0 < x < 3 2 0 x = 3 2 + x > 3

Therefore f(x) has a local maximum at x = 0 and a local minimum at x = 2/3.

of 3 MCV4U Sec.3.2 Note

Homework: p. # 1 , 2 , 4 , 5a , 7 , 8 , 11 , 12